GIFT 
Harry  Eas  t  Miller 


ROBINSON'S    MATHEMATICAL    SERIES. 


THE 


PROGRESSIVE 


HIGHER  ARITHMETIC, 


FOB 


SCHOOLS,  ACADEMIES,  AND  MERCANTILE  COLLEGES. 


FORMING  A  COMPLETE  TREATISE  ON  ARITHMETICAL  SCIENCE, 
AND  ITS  COMMERCIAL  AND  BUSINESS  APPLICATIONS. 


EDITED     BY 


DANIEL  W.    FISH,   A.M., 

EDITOR  or  ROBINSON'S  PROGRESSIVE  SERIES  OF  ARITHMETICS, 

AND  THE  SHORTER  COURSE. 


IVISON,  BLAKEMAN,  TAYLOR  &   CO., 
NEW  YORK  AND  CHICAGO. 

1878. 


. 

ROBINSON'S 

Mathematical   Series. 

Graded  to  the  wants  of  Primary,  Intermediate,  Grammar,  Normal, 
and  High  Schools,  Academies,  and  Colleges, 


Progressive  Table  Book. 
Progressive  Primary  Arithmetic. 
Progressive  Intellectual  Arithmetic. 
Rudiments  of  Written  Arithmetic. 

JUNIOR-CLASS  ARITHMETIC,  Oral  and  Written.    NEW. 

Progressive  Practical  Arithmetic. 

Key  to  Practical  Ai*ithrnetic. 
Progressive  Eiglier  Arithmetic. 

Key  to  Higher  Arithmetic, 
New  Elementary  Algebra. 

Key  to  Neif  Elementary  Algebra. 
New  University  Algebra. 

Key  to  New  University  Algebra. 
New  Geometry  and  Trigonometry.    In  one  vol. 
Geometry,  Plane  and  Solid.    In  separate  vol. 
Trigonometry,  Plane  and  Spherical.    In  separate  vol. 
New  Analytical  Geometry  and  Conic  Sections. 
New  Surveying  and  Navigation. 
New  Differential  and  Integral  Calculus. 
University  Astronomy — Descriptive  and  Physical. 

Key  to  Geometry  and  Trig.,  Analyt.  Geometry 
and  Conic  Sect.,  Surveying  and  Navigation. 


Copyright,  1860,  1863,  &  1875,  DANIEL  W.  FISH. 
Electrotyped  by  SMITH  &  McDouGAL,  82  Beekman  St.,N.  Y. 

GIFT  OF 


>«-«~*A 


PREFACE. 


THIS  work  is  intended  to  complete  a  well  graded  and 
progressive  series  of  Arithmetics,  and  to  furnish  to  ad- 
vanced students  a  more  full  and  comprehensive  text-book 
on  the  Science  of  Numbers  than  has  before  been  published , 
a  work  that  shall  embrace  those  subjects  necessary  to  give 
the  pupil  a  thoroughly  practical  and  scientific  arithmetical 
education,  either  for  the  farm,  the  workshop,  or  a  profes- 
sion, or  for  the  more  difficult  operations  of  the  counting- 
room  and  of  mercantile  and  commercial  life. 

There  are  two  general  methods  of  presenting  the  ele- 
ments of  arithmetical  science,  the  Synthetic  and  the  Ana- 
lytic. Comparison  enters  into  every  operation,  from  the 
simplest  combination  of  numbers  to  the  most  complicated 
problems  in  the  Higher  Mathematics.  Analysis  first 
generalizes  a  subject  and  then  develops  the  particulars  of 
which  it  consists;  Synthesis  first  presents  particulars, 
from  which,  by  easy  and  progressive  steps,  the  pupil  is  led 
to  a  general  and  comprehensive  view  of  the  subject. 
Analysis  separates  truths  and  properties  into  their  ele- 
ments or  first  principles;  Synthesis  constructs  general 
principles  from  particular  cases.  Analysis  appeals  more 
to  the  reason,  and  cultivates  the  desire  to  search  for  first 
principles,  and  to  understand  the  reason  for  every  process 
rather  than  to  know  the  rule.  Hence,  the  leading  method  in 
an  elementary  course  of  instruction  should  be  the  Synthetic, 
while  in  an  advanced  course  it  should  be  the  Analytic. 

The  following  characteristics  of  a  first  class  text-book 
will  be  obvious  to  all  who  examine  this  work :  the  typogra- 

M81911  (iii) 


iv  PREFACE. ' 

phy  and  mechanical  execution;  the  philosophical  and 
scientific  arrangement  of  the  subjects ;  dear  and  concise 
definitions ;  full  and  rigid  analyses ;  exact  and  compre- 
hensive rules ;  brief  and  accurate  methods  of  operation : 
the  wide  range  of  subjects  and  the  large  number  and  prac- 
tical character  of  the  examples — in  a  word,  SCIENTIFIC  AC- 
CUEACY  combined  with  PEACTICAL  UTILITY,  throughout  the 
entire  work. 

Much  labor  and  attention  have  been  devoted  to  obtain- 
ing correct  and  adequate  information  pertaining  to  mer- 
cantile and  commercial  transactions,  and  the  Government 
Standard  units  of  measures,  weights,  and  money.  The 
counting-room,  the  bank,  the  insurance  and  broker's  office, 
the  navy  and  ship-yard,  the  manufactory,  the  wharves,  the 
custom-house,  and  the  mint,  have  all  been  visited,  and  the 
most  reliable  statistics  and  the  latest  statutes  have  been 
consulted,  for  the  purpose  of  securing  entire  accuracy  in 
those  parts  of  this  work  which  relate  to  these  ;ubjects 
and  departments.  As  the  result  of  this  thorough  investi- 
gation, many  statements  found  in  most  other  arithmetics 
of  a  similar  grade  will  not  agree  with  the  facts  presented  in 
this  work,  and  simply  because  the  statements  in  these 
other  books  have  been  copied  from  older  works,  while  laws 
and  customs  have  undergone  great  changes  since  the  older 
works  were  written. 

New  material  and  new  methods  will  be  found  in  the  seve- 
ral subjects  throughout  the  entire  work.  Considerable  pro- 
minence has  been  given  to  Percentage  and  its  numerous  ap- 
plications, especially  to  Stocks,  Insurance,  Interest,  Aver- 
aging Accounts,  Domestic  and  Foreign  Exchange,  and  seve- 
ral other  subjects  necessary  to  quality  students  to  become 
good  accountants  or  commercial  business  men.  And  while 
this  work  may  embrace  many  subjects  not  necessary  to  the 


PREFACE.  V 

course  usually  prescribed  in  Mercantile  and  Commercial 
Colleges,  yet  those  subjects  requisite  to  make  good  account- 
ants, and  which  have  been  taught  orally  in  that  class  of 
institutions  from  want  of  a  suitable  text-book,  are  fully  dis- 
cussed and  practically  applied  in  this  work ;  and  it  is  there- 
fore believed  to  be  better  adapted  to  the  wants  of  Mercan- 
tile Colleges  than  any  similar  work  yet  published.  And 
while  it  is  due,  it  is  also  proper  here  to  state  that  J.  C. 
Porter,  A.  M.;  an  experienced  and  successful  teacher  of 
Mathematics  in  this  State,  and  formerly  professor  of  Com- 
mercial Arithmetic,  in  Iron  City  Commercial  College,  Pitts- 
burgh, Penn.,  has  rendered  valuable  aid  in  the  preparation 
of  the  above-named  subjects,  and  of  other  portions  of  the 
work.  He  is  likewise  the  author  of  the  Factor  Table  on 
pages  72  and  73,  and  of  the  new  and  valuable  improvement 
in  the  method  of  Cube  Root. 

Teachers  entertain  various  views  relative  to  having  the 
answers  to  problems  and  examples  inserted  in  a  text-book. 
Some  desire  the  answers  placed  immediately  after  the  ex- 
amples ;  others  wish  them  placed  together  in  the  back  part 
of  the  book;  and  still  others  desire  them  omitted  alto- 
gether. All  these  methods  have  their  advantages  and  their 
disadvantages. 

If  all  the  answers  are  given,  there  is  danger  that  the 
pupil  will  become  careless,  and  not  depend  enough  upon  the 
accuracy  of  his  own  computations.  Hence  he  is  liable  to 
neglect  the  cultivation  of  those  habits  of  patient  investiga- 
tion and  self-reliance  which  would  result  from  his  being 
obliged  to  test  the  truth  and  accuracy  of  his  own  processes f 
by  proof, — the  only  test  he  will  have  to  depend  upon  in  all 
the  computations  in  real  business  transactions  in  after  life. 
Besides,  the  work  of  proving  the  correctness  of  a  result  is 
often  of  quite  as  much  value  to  the  pupil  as  the  work  of 
1* 


^  PREFACE. 

performing  the  operation ;  as  the  proof  may  render  simple 
and  clear  some  part  or  the  whole  of  an  operation  that  was 
before  complicated  and  obscure. 

The  improvements  in  Percentage  made  necessary  by  the 
financial  changes  of  the  last  few  years  are  especially  notice- 
able. The  different  kinds  of  United  States'  Securities, 
Bonds,  and  Treasury  Notes  are  described,  and  their  com- 
parative value  in  commercial  transactions  illustrated  by 
practical  examples.  The  difference  between  Gold  and 
Currency,  and  the  corresponding  difference  in  prices,  ex- 
hibited in  trade,  are  taught  and  illustrated,  and  many 
other  things  that  every  commercial  student  and  business 
man  ought  to  know  and  understand. 
AUGUST,  1860. 


IMPROVED  EDITION. 

Such  changes  only  have  been  made  in  the  present  edition  as  were 
necessary  to  conform  to  law  and  usage,  and  to  meet  a  demand  from 
many  of  the  best  teachers  of  the  country  for  a  full  and  practical  treatise 
on  Mensuration.  Hence  Foreign  Exchange  has  been  so  modified  in  the 
Tables  and  Examples  as  to  conform  to  the  Act  of  March  3,  1873,  and  to 
present  usage. 

Thirty -six  pages  of  useful  and  practical  matter  on  Mensuration  and 
Measurements,  have  been  carefully  prepared  and  substituted  at  the  end 
of  the  book  for  the  lengthy  treatise  of  the  Metric  System  heretofore 
presented,  and  which  is  scarcely  ever  used  in  this  country. 

To  avoid  repetition,  as  well  as  to  put  in  a  more  condensed  form,  the 
Principles  and  Applications  of  the  Square  and  Cube  roots  that  inter- 
vened between  "Evolution"  and  "Series"  in  former  editions,  have 
been  embodied  in  these  thirty-six  pages,  and  also  so  much  of  the 
Metric  System  as  is  of  any  practical  value. 

It  is  hoped  that  these  improvements  will  give  new  life  to  a  book  that 
has  already  proved  its  merits  by  the  large  and  increasing  circulation 
it  has  obtained. 

JULY,  1875, 


CONTENTS. 


PAGE 

Definitions 11 

Signs 13 

Axioms , 14 

Notation  and  Numeration 15 

SIMPLE    NUMBERS. 

Addition ! 23 

Adding  two  or  more  columns  at  one  operation « 27 

Subtraction ...„ 30 

Two  or  more  subtrahends 33 

Multiplication 35 

Powers  of  Numbers 39 

Continued  Multiplication 40 

Contractions  in  Multiplication... 41 

Division , 47 

Abbreviated  Long  Division 50 

Successive  Division 55 

Contractions  in  Division , 55 

General  Problems  in  Simple  Numbers  61 

PROPERTIES   OF   NUMBERS. 

Kxact  Divisors 65 

Prime  Numbers 68 

Table  of  Prime  Numbers , 70 

Factoring 70 

Factor  Table .  72 

Greatest  Common  Divisor 76 

Least  Common  Multiple 82 

Cancellation  : * 86 

FRACTIONS. 

Definitions,  Notation  and  Numeration 89 

Reduction , 92 

Addition 99 

(Vii) 


Viii  CONTENTS. 

PAQB 

Subtraction « 101 

Theory  of  Multiplication  and  Division 103 

Multiplication 104 

Division 107 

Greatest  Common  Divisor Ill 

Least  Common  Multiple , 112 

DECIMALS. 

Notation  and  Numeration 117 

Reduction 121 

Addition 124 

Subtraction 126 

Multiplication « 127 

Contracted  Multiplication 128 

Division 132 

Contracted  Division 134 

Circulating  Decimals 136 

Reduction  of  Circulating  Decimals 139 

Addition  and  Subtraction  of  Circulating  Decimals 142 

Multiplication  and  Division  of  Circulating  Decimals 144 

UNITED    STATES    MONEY. 

Notation  and  Numeration 145 

Reduction 147 

Operations 147 

Problems «... , 150 

Ledger  Accounts >  153 

Accounts  and  Bills 153 

Continued  Fractions 161 

COMPOUND    NUMBERS. 

Measures  of  Extension 164 

Measures  of  Capacity 170 

Measures  of  Weight 171 

Measure  of  Time 175 

Measure  of  Angles 177 

Miscellaneous  Tables , 178 

Government  Standards  of  Measures  and  Weights 179 

English  Measures  and  Weights 182 

French  Measures  and  Weights I84 

Money  and  Currencies 1$*? 

Reduction 192 

Reduction  Descending 192 


CONTENTS.  'ix 

PACK 

Reduction  Ascending , 199 

Addition 206 

Subtraction 209 

Multiplication 214 

Division 216 

Longitude  and  Time 218 

DUODECIMALS. 

Addition  and  Subtraction 227 

Multiplication 228 

Division 230 

SHORT    METHODS. 

For  Subtraction 232 

For  Multiplication 233 

For  Division 241 

RATIO.  243 

PROPORTION.  247 

Cause  and  Effect ,.    » ,  249 

Single  Proportion 249 

Compound  Proportion , 253 

PERCENTAGE. 

Notation 259 

General  Problems - 260 

Applications 268 

Commission 268 

Stocks , 272 

Stock-jobbing 273 

Instalments,  Assessments,  and  Dividends 276 

Stock  Investments 279 

Gold  Investments 285 

Profit  and  Loss 287 

Insurance 291 

Life  Insurance 293 

Life  Table 295 

Endowment  Assurance  Table 296 

Taxes 298 

General  Average 301 

Custom  House  Business 303 

Simple  Interest 307 


X  CONTENTS. 

4 

PASS 

Partial  Payments  or  Indorsements 314 

Savings  Banks  Accounts 319 

Compound  Interest 321 

Discount 328 

Banking 330 

Exchange 337 

Direct  Exchange 339 

Table  of  Foreign  Coins  and  Money 342 

Arbitrated  Exchange 348 

Equation  of  Payments 352 

Compound  Equations •• 357 

Partnership 364 

ALLIGATION 370 

INVOLUTION 379 


EVOLUTION. 


Square  Root , 

Contracted  Method. 

Cube  Root 

Contracted  Method. 


SERIES.  398 

Arithmetical  Progression 400 

Geometrical  Progression 403 

Annuities 408 

MISCELLANEOUS  EXAMPLES 414 

MENSURATION. 

Lines  and  Angles 421 

Triangles 422 

Quadrilaterals 426 

Circles 428 

Similar  Plane  Figures 432 

Solids 435 

Prisms  and  Cylinders 430 

Pyramids  and  Cones 437 

Spheres 438 

Similar  Solids. 441 

Gauging 443 

Measurement  of  Land 445 

Boards  and  Timber 448 

Masonry 449 

Capacity  of  Bins.  Cisterns,  etc 450 

METRIC  SYSTEM..                                                                                                   ..  453 


HIGHER  ARITHMETIC. 


DEFINITIONS. 

I.  Quantity  is  any  thing  that  can  be  increased,  diminished,  or 
measured ;  as  distance,  space,  weight,  motion,  time. 

<J.    A  Unit  is  one,  a  single, thing,  or  a  definite  quantity. 

3.  A  Number  is  a  unit,  or  a  collection  of  units. 

4.  The  Unit  of  a  Number  is  one  of  the  collection  constituting 
the  number.     Thus,  the  unit  of  34  is  1 ;  of  34  days  is  1  day. 

5.  An  Abstract  Number  is  a  number  used  without  reference 
to  any  particular  thing  or  quantity;  as  3,  24,  756. 

O.   A  Concrete  Number  is  a  number  used  with  reference  to 
some  particular  thing  or  quantity;  as  21  hours,  4  cents,  230  miles. 
T'.   Unity  is  the  unit  of  an  abstract  number. 

8.  The  Denomination  is  the  name  of  the  unit  of  a  concrete 
number. 

9.  A  Simple  Number  is  either  an  abstract  number,  or  a  con- 
crete number  of  but  one  denomination;  as  48,  52  pounds,  36  days. 

10.  A  Compound  Number  is  a  concrete  number  expressed  in 
two  or  more  denominations ;  as,  4  bushels  3  pecks,  8  rods  4  yards 
2  feet  3  inches. 

II.  An  Integral  Number,  or  Integer,  is  a  number  which  ex- 
presses whole  things ;  as  5,  12  dollars,  17  men. 

12.  A  Fractional  Number,  or  Fraction,  is  a  number  which 
expresses  equal  parts  of  a  whole  thing  or  quantity;  as  &,  f  of  a 
pound,  y^  of  a  bushel. 

13.  Like  Numbers  have  the  same  kind  of  unit,  or  express  the 
same  kind  of  quantity.     Thus,  74  and  16  are  like  numbers;  so 
are  74  pounds,  16  pounds,  and  12  pounds;  also,  4  weeks  3  days,  and 
16  minutes  20  seconds,  both  being  used  to  express  units  of  time. 

14.  Unlike  Numbers  have  different  kinds  of  units,  or  are  used 


12  SIMPLE  NUMBERS. 

i 

to  Express  ,di,fier,e?,c;j  kinds  of  quantity.     Thus,  36  miles,  and  15 
'  days  |»  5>  hours,  36  minutes,  and  7  bushels  3  pecks. 
-    15.  '"A  Power  ?.p  the  product  arising  from  multiplying  a  number 
by  itself,  or  repeating  it  any  number  of  times  as  a  factor. 
1G.   A  Root  is  a  factor  repeated  to  produce  a  power. 

17.  A  Scale  is  the  order  of  progression  on  which  any  system 
of  notation  is  founded.     Scales  are  uniform  and  varying. 

18.  A  Uniform  Scale  is  one  in  which  the  order  of  progression 
is  the  same  throughout  the  entire  succession  of  units. 

1O.  A  Varying  Scale  is  one  in  which  the  order  of  progression 
is  not  the  same  throughout  the  entire  succession  of  units. 

20.  A  Decimal  Scale  is  one  in  which  the  order  of  progression 
is  uniformly  ten. 

21.  Mathematics  is  the  science  of  quantity. 

The  two  fundamental  branches  of  Mathematics  are  Geometry 
and  Arithmetic.  Geometry  considers  quantity  with  reference  to 
positions,  form,  and  extension.  Arithmetic  considers  quantity  as 
an  assemblage  of  definite  portions,  and  treats  only  of  those  condi- 
tions and  attributes  which  may  be  investigated  and  expressed  by 
numbers.  Hence, 

22.  Arithmetic  is  the  Science  of  numbers,  and  the  Art  of 
computation. 

NOTE. — When  Arithmetic  treats  of  operations  on  abstract  numbers  it  is  a  sci- 
ence, and  is  then  called  Pure  Arithmetic.  When  it  treats  of  operations  on  con- 
crete numbers  it  is  an  art,  and  is  then  called  Applied  Arithmetic.  Pure  and 
Applied  Arithmetic  are  also  called  Theoretical  and  Practical  Arithmetic. 

23.  A  Demonstration  is  a  process  of  reasoning  by  which  a 
truth  or  principle  is  established. 

24.  An  Operation  is  a  process  in  which  figures  are  employed 
to  make  a  computation,  or  obtain  some  arithmetical  result. 

25.  A  Problem  is  a  question  requiring  an  operation. 

26.  A  Rule  is  a  prescribed  method  of  performing  an  operation. 

27.  Analysis,  in  arithmetic,  is  the  process  of  investigating 
principles,  and  solving  problems,  independently  of  set  rules. 

28.  The  Five  Fundamental  Operations  of  Arithmetic  are, 
Notation  and  Numeration,  Addition,  Subtraction,  Multiplication, 
and  Division. 


DEFINITIONS.  13 

SIGNS. 

29.  A  Sign  is  a  character  indicating  the  relation  of  numbers, 
or  an  operation  to  be  performed. 

30.  The  Sign  of  Numeration  is  the  comma  (,).     It  indicates 
that  the  figures  set  off  by  it  express  units  of  the  same  general  name, 
and  are  to  be  read  together,  as  thousands,  millions,  billions,  etc. 

31.  The  Decimal  Sign  is  the  period  (.).     It  indicates  that 
the  number  after  it  is  a  decimal. 

32.  The  Sign  of  Addition  is  the  perpendicular  cross,  -f,  called 
plus.     It  indicates  that  the  numbers  connected  by  it  are  to  be 
added ;  as  3  -f  5  +  7,  read  3  plus  5  plus  7. 

33.  The  Sign  of  Subtraction  is  a  short  horizontal  line,  — , 
called  minus.     It  indicates  that  the  number  after  it  is  to  be  sub- 
tracted from  the  number  before  it;  as  12  —  7,  read  12  minus  7. 

34.  The  Sign  of  Multiplication  is  the  oblique  cross,  x  .     It 
indicates  that  the  numbers  connected  by  it  are  to  be  multiplied 
together;  as  5  x  3  x  9,  read  5  multiplied  by  3  multiplied  by  9. 

35.  The  Sign  of  Division  is  a  short  horizontal  line,  with  a 
point  above  and  one  below,  -5-.      It  indicates  that  the  number 
before  it  is  to  be  divided  by  the  number  after  it ;  as  18  -f-  6,  read 
18  divided  by  6. 

Division  is  also  expressed  by  writing  the  dividend  above,  and 
the  divisor  below,  a  short'  horizontal  line.  Thus,  *g8,  read  18 
divided  by  6. 

36.  The  Sign  of  Equality  is  two  short,  parallel,  horizontal 
lines,  =.      It  indicates  that  the  numbers,  or  combinations  of 
numbers,  connected  by  it  are  equal;  as  4  -f  8  =  15  —  3,  read  4 
plus  8  is  equal  to  15  minus  3.     Expressions  connected  by  the 
sign  of  equality  are  called  equations. 

37.  The  Sign  of  Aggregation  is  a  parenthesis,  (  ).     It  indi- 
cates that  the  numbers  included  within  it  are  to  be  considered 
together,  and  subjected  to  the  same  operation.     Thus,  (8  -f  4)  X  5 
indicates  that  both  8  and  4,  or  their  sum,  is  to  be  multiplied  by  5. 

A  vinculum  or  bar,  ,  has  the  same  signification.     Thus, 

7x  9-v-3  =  21. 


14  SIMPLE  NUMBERS. 

38.  The  Sign  of  Ratio  is  two  points,  :  .     Thus,  7  :  4  is  read, 
the  ratio  of  7  to  4. 

39.  The  Sign  of  Proportion  is   four  points,   :  :  .     Thus, 
3  :  6  :  :  4  :  8,  is  read,  3  is  to  6  as  4  is  to  8. 

40.  The  Sign  of  Involution  is  a  number  written  above,  and  a 
little  to  the  right,  of  another  number.     It  indicates  the  power  to 
which  the  latter  is  to  be  raised.     Thus,  12s  indicates  that  12  is 
to  be  taken  3  times  as  a  factor;  the  expression  is  equivalent  to 
12  x  12  x  12.     The  number  expressing  the  sign  of  involution  is 
called  the  Index  or  Exponent. 

41.  The  Sign  of  Evolution,  -y/,  is  a  modification  of  the  letter  r. 
It  indicates  that  some  root  of  the  number  after  it  is  to  be  extracted. 
Thus,  v/25  indicates  that  the  square  root  of  25  is  to  be  extracted; 

indicates  that  the  cube  root  of  64  is  to  be  extracted. 


AXIOMS. 

An  Axiom  is  a  self-evident  truth.     The  principal  axioms 
required  in  arithmetical  investigations  are  the  following  : 

1.  If  the  same  quantity  or  equal  quantities  be  added  to  equal 
quantities,  the  sums  will  be  equal. 

2.  If  the  same  quantity  or  equal  quantities  be  subtracted  from 
equal  quantities,  the  remainders  will  be  equal. 

3.  If  equal  quantities  be  multiplied  by  the  same  number,  the 
products  will  be  equal. 

4.  If  equal  quantities  be  divided  by  the  same  number,  the  quo- 
tients will  be  equal. 

5.  If  the  same  number  be  added  to  a  quantity  and  subtracted 
from  the  sum,  the  remainder  will  be  that  quantity. 

6.  If  a  quantity  be  multiplied  by  a  number  and  the  product 
divided  by  the  same  number,  the  quotient  will  be  that  quantity. 

7.  Quantities  which  are  respectively  equal  to  any  other  quantity 
are  equal  to  each  other. 

8.  Like  powers  or  like  roots  of  equal  quantities  are  equal. 

9    The  whole  of  any  quantity  is  greater  than  any  of  its  parts. 
10.  The  whole  of  any  quantity  is  equal  to  the  sum  of  all  its 
parts. 


NOTATION  AND  NUMERATION.  15 


NOTATION  AND  NUMERATION. 

4:3.  Notation  is  a  system  of  writing  or  expressing  numbers  by 
characters;  and, 

44.  Numeration  is  a  method  of  reading  numbers  expressed 
by  characters. 

45.  Two  systems  of  notation  are  in  general  use  —  the  Roman 
and  the  Arabic. 

NOTE.  —  The  Roman  Notation  is  supposed  to  have  been  first  used  by  the 
Romans  :  hence  its  name.  The  Arabic  Notation  was  first  introduced  into  Europe 
by  the  Moors  or  Arabs,  who  conquered  and  held  possession  of  Spain  during  the 
llth  century.  It  received  the  attention  of  scientific  men  in  Italy  at  the  begin- 
ning of  the  13th  century,  and  was  soon  afterward  adopted  in  most  European 
countries.  Formerly  it  was  supposed  to  be  an  invention  of  the  Arabs;  but 
investigations  have  shown  that  the  Arabs  adopted  it  from  the  Hindoos,  among 
whom  it  has  been  in  use  more  than  2000  years.  From  this  undoubted  origin  it 
is  sometimes  called  the  Indian  Notation. 

THE   ROMAN    NOTATION. 

4G.  Employs  seven  capital  letters  to  express  numbers.  Thus, 
Letters,  I  V  X  L  C  D  M 

Values,  one,          five,  ten.  fifty,      ,    °°e  J     ,_   fi!e  „       ,    one 

Jj       hundred,    hundred,    thousand. 

47.  The  Roman  notation  is  founded  upon  five  principles,  as 
follows : 

1st.  Repeating  a  letter  repeats  its  value.  Thus,  II  represents 
two,  XX  twenty,  CCC  three  hundred. 

2d.  If  a  letter  of  any  value  be  placed  after  one  of  greater  value, 
its  value  is  to  be  united  to  that  of  the  greater.  Thus,  XI  repre- 
sents eleven,  LX  sixty,  DC  six  hundred. 

3d.  If  a  letter  of  any  value  be  placed  before  one  of  greater  value, 
its  value  is  to  be  taken  from  that  of  the  greater.  Thus,  IX  repre- 
sents nine,  XL  forty,  CD  four  hundred. 

4th.  If  a  letter  of  any  value  be  placed  between  two  letters,  each 
of  greater  value,  its  value  is  to  be  taken  from  the  united  value  of 
the  other  two.  Thus,  XIY  represents  fourteen,  XXIX  twenty- 
nine,  XCIY  ninety-four. 

5th.  A  bar  or  dash  placed  over  a  letter  increases  its  value  one 
thousand  fold.  Thus,  V  signifies  five,  and  V  five  thousand ;  L 
fifty,  and  L  fifty  thousand. 


16  SIMPLE  NUMBERS. 

i 

TABLE    OF   ROMAN    NOTATION. 

I  is  One.  XX  is  Twenty. 

II  "  Two.  XXI  "  Twenty-one. 

III  "  Three.  XXX  "  Thirty. 

IV  "  Four.  XL  "  Forty. 
V  "  Five.  L  "  Fifty. 

VI  "  Six.  LX  "  Sixty. 

VII'"  Seven.  LXX  "  Seventy. 

VIII  "  Eight.  LXXX  "  Eighty. 

IX  "  Nine.  XC  "  Ninety. 

X  "  Ten.  C   "  One  hundred. 

XI  "  Eleven.  CO  "  Two  hundred. 

XII  "  Twelve.  D  "  Five  hundred. 

XIII  "  Thirteen.  DC  "  Six  hundred. 

£IV  "  Fourteen.  M  "  One  thousand.       [dred. 

XV  "  Fifteen.  MC  "  One  thousand  one  hun- 

XVI  "  Sixteen.  MM  "  Two  thousand. 

XVII  "  Seventeen.  X  "  Ten  thousand. 

XVIII  "  Eighteen.  _C  "  One  hundred  thousand. 

XIX  "  Nineteen.  M  "  One  million. 

NOTES. — 1.  Though  the  letters  used  in  the  above  table  have  been  employed 
nr  the  Roman  numerals  for  many  centuries,  the  marks  or  characters  used  origi- 
nally in  this  notation  are  as  follows : 

Modern  numerals,  I         V         X         L         C         D         M 

Primitive  characters,          I          V          X          L         C          N          M 

2.  The  system  of  Roman  Notation  is  not  welt  adapted  to  the  purposes  of  nu- 
merical calculation ;  it  is  principally  confined  to  the  numbering  of  chapters  and 
«ections  of  books,  public  documents,  etc. 

EXAMPLES    FOR    PRACTICE. 

Express  the  following  numbers  by  the  Roman  notation: 

1.  Fourteen.  6.  Fifty-one. 

2.  Nineteen.  7.  Eighty-eight. 

3.  Twenty-four.  8  Seventy-three. 

4.  Thirty-nine.  9.  Ninety-five. 

5.  Forty-six.  10.  One  hundred  one. 

11.  Five  hundred  fifty-five. 

12.  Seven  hundred  ninety-eight. 

13.  One  thousand  three. 

14.  Twenty  thousand  eight  hundred  forty-five. 


NOTATION  AND  NUMERATION.  If 

THE   ARABIC    NOTATION 

48.    Employs  ten  characters  or  figures  to  express  numbers. 
Thus, 
Figures,  .0         123456789 

Names  and )     nau8ht      one>      two»     three,    four,     five,       six,     seven,    eight,    nine, 
cipher. 


values.  or 


49.  The  cipher,  or  first  character,  is  called  naught,  because  it 
has  no  value  of  its  own.     It  is  otherwise  termed  nothing,  and  zero. 
The  other  nine  characters  are  called  significant  figures,  because 
each  has  a  value  of  its  own.     They  are  also  called  digits,  a  word 
derived  from  the  Latin  term  diyitus,  which  signifies  finger. 

50.  The  ten  Arabic  characters  are  the  Alphabet  of  Arithmetic. 
Used  independently,  they  can  express  only  the  nine  numbers  that 
correspond  to  the  names  of  the  nine  digits.     But  when  combined 
according  to  certain  principles,  they  serve  to  express  all  numbers. 

01.  The  notation  of  all  numbers  by  the  ten  figures  is  accom- 
plished by  the  formation  of  a  series  of  units  of  different  values,  to 
which  the  digits  may  be  successively  applied.  First  ten  simple 
units  are  considered  together,  and  treated  as  a  single  superior 
unit;  then,  a  collection  often  of  these  new  units  is  taken  as  a  still 
higher  unit;  and  so  on,  indefinitely.  A  regular  series  of  units,  in 
ascending  orders,  is  thus  formed,  as  shown  in  the  following 

TABLE    OF    UNITS 

Primary  units  are  called  units  of  the  first    order. 

Ten  units  of  the  first      order  make  1  unit    "     "    second    " 
Ten     "      "     "    second     "         "      1     "      "     "    third       " 
Ten     "      "     "    third        "         "      1     "      "     "    fourth     " 
etc.,        etc.  etc.,         etc. 


The  various  orders  of  units,  when  expressed  by  figures, 
are  distinguished  from  each  other  by  their  location,  or  the  place 
they  occupy  in  a  horizontal  row  of  figures.  Units  of  the  first  order 
are  written  at  the  right  hand  ;  units  of  the  second  order  occupy 
the  second  place  ;  units  of  the  third  order  the  third  place  ;  and  so 
on,  counting  from  right  to  left,  as  shown  on  the  following  page  : 


SIMPLE  NUMBERS. 


000000000 
In  this  notation  we  observe  — 

1st.  That  a  figure  written  in  the  place  of  any  order,  expresses 
as  many  units  of  that  order  as  is  denoted  by  the  name  of  the  figure 
used.  Thus,  436  expresses  4  units  of  the  3d  order,  3  units  of  the 
2d  order,  and  6  units  of  the  1st  order. 

2d.  The  cipher,  having  no  value  of  its  own,  is  used  to  fill  the 
places  of  vacant  orders,  and  thus  preserve  the  relative  positions  of 
the  significant  figures.  Thus,  in  50,  the  cipher  shows  the  absence 
of  simple  units,  and  at  the  same  time  gives  to  the  figure  5  the 
local  value  of  the  second  order  of  units. 

54.  Since  the  number  expressed  by  any  figure  depends  upon 
the   place   it  occupies,  it  follows   that  figures  have  two  values, 
Simple  and  Local. 

55.  The  Simple  Value  of  a  figure  is  its  value  when  taken 
alone  ;  thus,  4,  7,  2. 

5G.  The  Local  Value  of  a  figure  is  its  value  when  used  with 
another  figure  or  figures  in  the  same  number.  Thus,  in  325,  the 
local  value  of  the  3  is  300,  of  the  2  is  20,  and  of  the  5  is  5  units. 

NOTE.  —  When  a  figure  occupies  units'  place,  its  simple  and  local  values  are 
the  same. 

57.  The  leading  principles  upon  which  the  Arabic  notation 
is  founded  are  embraced  in  the  following 

GENERAL   LAWS. 

I.  AH  numbers  are  expressed  by  applying  the  ten  figures  to  dif- 
ferent orders  of  units. 

II.  The  different  orders  of  units  increase  from  right  to  left,  and 
decrease  from  left  to  right,  in  a  tenfold  ratio. 

III.  Every  removal  of  a  figure  one  place  to  the  left,  increases  its 
local  value  tenfold;  and  every  removal  of  a  figure  one  place  to  the 
right,  diminishes  its  local  value  tenfold. 


NOTATION  AND  NUMERATION.  19 

08.  In  numerating,  or  expressing  numbers  verbally,  the  various 
orders  of  units  have  the  following  names : 

ORDERS.  NAMES. 

1st  order  is  called  Units. 

2d    order  "  "  Tens. 

3d    order  "  "  Hundreds. 

4th  order  "  "  Thousands. 

5th  order  "  "  Tens  of  thousands. 

6th  order  "  "  Hundreds  of  thousands. 

7th  order  "  "  Millions. 

8th  order  "  "  Tens  of  millions. 

9th  order  "  "  Hundreds  of  millions, 

etc.,  etc.  etc.,     etc. 

50.  This  method  of  numerating,  or  naming,  groups  the  suc- 
cessive orders  into  period*  of  three  figures  each,  there  being  three 
orders  of  thousands,  three  orders  of  millions,  and  so  on  in  all 
higher  orders.  These  periods  are  commonly  separated  by  commas, 
as  in  the  following  table,  which  gives  the  names'  of  the  orders 
and  periods  to  the  twenty-seventh  place. 

»  3 

a  no  fl  5  g£ 

M  M  .2  .  t  *9 

o          o  "-1  —  - 


w 

•3 

P-. 
8 

tw 

W 

3 

'-3 

M 

<D 
CM 

~d 

"3 

a4 

CM 

1 

CM 

§ 

43 

CM 

00 

a 

o 

3 

CM 

a 
.2 

'§ 

1 

.•§ 
'3 
p 

CM 

98, 7  65, 4  32, 109, 87  6, 5  56, 789, 012, 3  45 

ninth    eijrhth     seventh      sixth        fifth       fourth       third       second        first 
period,  period,      period,     period,    period,     period,      period,     period.      period. 

NOTE.  —  This  is  the  French  method  of  numerating,  and  is  the  one  in  general 
use  in  this  country.  The  English  numerate  hy  periods  of  six  figures  each. 

6O.  The  names  of  the  periods  are  derived  from  the  Latin 
numerals.  The  twenty-two  given  on  the  following  page  extend 
the  numeration  table  to  the  sixty-sixth  place  or  order,  inclusive. 


NAMES. 

PERIODS. 

NAMES. 

Units. 

12th 

Decillions. 

Thousands. 

13th 

Undecillions. 

Millions. 

14th 

Duodecillions 

Billions. 

15th 

Tredecillions. 

Trillions. 

16th 

Quatuordecillions. 

Quadrillions. 

17th 

Quindecillions. 

Quintillions. 

18th 

Sexdecillions. 

Sextillions. 

19th 

Septendecillions. 

Septillions. 

20th 

Octodecillions. 

Octillions. 

21st 

Novendecillions. 

Nonillions. 

22d 

Vigintillions. 

20  SIMPLE  NUMBERS. 

PERIODS. 

1st 

2d 

3d 

4th 

5th 

6th 

7th 

8th 

9th 
10th 
tlth 

61.    From  this  analysis  of  the  principles  of  Notation  and  Nume- 
ration, we  derive  the  following  rules : 

RULE   FOR   NOTATION. 

I.  Beginning  at  the  left  hand,  write  the  figures  belonging  to  the 
highest  period. 

II.  Write  the  hundreds,  tens,  and  units  of  each  successive  period 
in  their  order,  placing  a  cipher  wherever  an  order  of  units  is 
omitted. 

RULE   FOR   NUMERATION. 

1.  Separate  the  number  into  periods  of  three  figures  each,  com- 
mencing at  the  right  hand. 

II.  Beginning  at  the  left  hand,  read  each  period  separately,  and 
give  the  name  to  each  period,  except  the  last,  or  period  of  units. 
NOTE. — Omit  and  in  reading  the  orders  of  units  and  periods  of  a  number. 

EXAMPLES    FOR   PRACTICE. 

Write  and  read  the  following  numbers  :  — 
1    One  unit  of  the  3d  order,  two  of  the  2d,  five  of  the  1st. 
Ans.  125  ',  read,  one  hundred  twenty-Jive. 

2.  Two  units  of  the  5th  order,  four  of  the  4th,  five  of  the  2d, 
six  of  the  1st.     Ans.  24056;  read,  twenty-four  thousand  fiffy-six. 

3.  Seven  units  of  the  4th  order,  five  of  the  3d,  three  of  the  2d, 
eight  of  the  1st. 


NOTATION  AND  NUMERATION.  21 

4.  Nine  units  of  the  4th  order,  two  of  the  3d',  four  of  the  1st. 

5.  Five  units  of  the  4th  order,  eight  of  the  2d. 

6.  Five  units  of  the  5th  order,  one  of  the  3d,  eight  of  the  1st. 

7.  Three  units  of  the  5th  order,  six  of  the  4th,  four  of  the  3d, 
seven  of  the  1st. 

8.  Two  units  of  the  6th  order,  four  of  the  5th,  nine  of  the  4th, 
three  of  the  3d,  five  of  the  1st. 

9.  Three  units  of  the  8th  order,  five  of  the  7th,  four  of  the  6th, 
three  of  the  5th,  eight  of  the  4th,  five  of  the  3d,  eight  of  the  2d, 
seven  of  the  1st. 

10.  Three  units  of  the  9th  order,  eight  of  the  7th,  four  of  the 
6th,  six  of  the  5th,  nine  of  the  1st. 

11.  Five  units  of  the  12th  order,  three  of  the  llth,  six  of  the 
10th. 

12.  Four  units  of  the  12th  order,  five  of  the  10th,  eight  of  the 
5th,  nine  of  the  4th,  four  of  the  3d. 

13.  Three  units  of  the  15th  order,  six  of  the  14th,  five  of  the 
13th,  three  of  the  9th,  six  of  the  8th,  five  of  the  7th,  three  of  the 
3d,  six  of  the  2d,  five  of  the  1st. 

14.  Five  units  of  the  18th  order,  three  of  the  17th,  six  of  the 
16th,  four  of  the  15th,  seven  of  the  14th,  eight  of  the  13th,  four 
of  the  12th,  five  of  the  llth,  six  of  the  10th,  seven  of  the  9th, 
eight  of  the  8th,  nine  of  the  7th,  five  ot  the  6th,  six  of  the  5th, 
three  of  the  4th,  two  of  the  3d,  four  of  the  2d,  eight  of  the  1st. 

15.  Two  units  of  the  20th  order,  seven  of  the  19th,  four  of  the 
18th,  eight  of  the  13th,  five  of  the  6th,  five  of  the  5th;  five  of  the 
4th,  nine  of  the  1st. 

Write  the  following  numbers  in  figures : 

16.  Forty-eight. 

17.  One  hundred  sixty-four. 

18.  Forty-eight  thousand  seven  hundred  eighty-nine. 

19.  Five  hundred  thirty-six  million  three  hundred  forty-seven 
thousand  nine  hundred  seventy-two. 

20.  Ninety-nine  billion  thirty-seven  thousand  four. 

21.  Eight  hundred  sixty-four  billion  five  hundred  thirty-eight 
million  two  hundred  seventeen  thousand  nine  hundred  fifty-three. 


22  SIMPLE  NUMBERS. 

22.  One  hundred  seventeen  quadrillion  two  hundred  thirty-five 
trillion  one  hundred  four  billion  seven  hundred  fifty  million  sixty- 
six  thousand  ten. 

23.  Ninety-nine   quintillion   seven   hundred   forty-one   trillion 
fifty-four  billion  one  hundred  eleven  million  one  hundred  one. 

24.  One  hundred  octillion  one  hundred  septillion  one  hundred 
quintillion  one  hundred  quadrillion  one  hundred  trillion  one  hundred 
billion  one  hundred  million  one  hundred  thousand  one  hundred. 

25.  Four  decillion   seventy-five  nonillion  three  octillion  fifty- 
two  septillion  one  sextillion  four  hundred  seventeen  quintillion 
ten  quadrillion  twelve  trillion  fourteen  billion  three  hundred  sixty 
million  twenty-two  thousand  five  hundred  nineteen. 

Write  the  following  numbers  in  figures,  and  read  them : 

26.  Twenty-five  units  in  the  2d  period,  four  hundred  ninety-six 
in  the  1st.  Ans.  25,496. 

27.  Three  hundred  sixty-four  units  in  the  3d   period,  seven 
hundred  fifteen  in  the  2d,  eight  hundred  thirty-two  in  the  1st. 

28.  Four  hundred  thirty-six  units  in  the  4th  period,  twelve  in 
the  3d,  one  hundred  in  the  2d,  three  hundred  one  in  the  1st. 

29.  Eighty-one  units  in  the  5th  period,  two  hundred  nineteen 
in  the  4th,  fifty-six  in  the  2d. 

30.  Nine  hundred  forty-five  units  in  the  7th  period,  eighteen  in 
the  5th,  one  hundred  three  in  the  3d. 

31.  One  unit  in  the  10th  period,  five  hundred  thirty-six  in  the 
9th,  two  hundred  forty-seven  in  the  8th,  nine  hundred  twenty-four 
in  the  7th. 

Point  off  and  read  the  following  numbers : 


32.  564. 

33.  24835. 

34.  2474783. 

35.  247843112. 

36.  23678542789. 


37.  2005. 

38.  100103. 

39.  53000008. 

40.  1001005003. 

41.  750000040003. 


42.  247364582327896438542721. 

43.  379403270506038009503070. 

44.  20005700032004673000430512500000567304705030040. 


ADDITION.  23 


ADDITION. 

Addition  is  the  process  of  uniting  several  numbers  of  the 
same  kind  into  one  equivalent  number. 

63.  The  Sum  or  Amount  is  the  result  obtained  by  the  process 
•?f  addition. 

64.  When  the  given  numbers  contain  several  orders  of  units, 
the  method  of  addition  is  based  upon  the  following  principles : 

I.  If  the  like  orders  of  units  be  added  separately,  the  sum  of 
all  the  results  must  be  equal  to  the  entire  sum  of  the  given  num- 
bers.   (Ax.  10). 

II.  If  the  sum  of  the  units  of  any  order  contain  units  of  a 
higher  order,  these  higher  units  must  be  combined  with  units  of 
like  order.     Hence, 

III.  The  work  must  commence  with  the  lowest  unit,  in  ordei 
to  combine  the  partial  sums  in  a  single  expression,  at  one  ope- 
ration. 

I.  Find  the  sum  of  397,  476,  and  873. 

OPERATION  ANALYSIS.      We   arrange   the   numbers   so  that 

397  units  of  like  order  shall  stand  in  the  same  column. 

We  then  add  the  first,  or  right  hand  column,  and 

find  the  sum  to  be  16  units,  or  1  ten  and  6  units ; 

1746  writing  the  6  units  under  the  column  of  units,  we 

add  the  1  ten  to  the  column  of  tens,  and  find  the 

sum  to  be  24  tens,  or  2  hundreds  and  4  tens  ;  writing  the  4  tens  under 

the  column  of  tens,  we  add  the  2  hundreds  to  the  column  of  hundreds, 

and  find  the  sum  to  be  17  hundreds,  or  1  thousand  and  7  hundreds ; 

writing  the  7  hundreds  under  the  column  of  hundreds,  and  the  1  in 

thousands'  place,  we  have  the  entire  sum,  1746. 

65.    From  these  principles  we  deduce  the  following 

RULE.     I.   Write  the  numbers  to  be  added  so  that  all  the  unit* 

of  the  same  order  shall  stand  in  the  same  column  ;   that  is,  units 

under  units,  tens  under  tens,  etc. 

II.  Commencing  at  units,  add  each  column  separately,  and  write 
the  sum  underneath,  if  it  be  less  than  ten. 


24  SIMPLE  NUMBERS. 

III.  If  the  sum  of  any  column  be  ten  or  more  than  ten,  write  the 
unit  figure  only,  and  add  the  ten  or  tens  to  the  next  column. 

IV.  Write  the  entire  sum  of  the  last  column. 

NOTES. — 1.  In  adding,  learn  to  pronounce  the  partial  results  without  naming 
the  figures  separately.  Thus,  in  the  operation  given  for  illustration,  say  3,  9, 
16;  8,  15,  24;  10,  14,  17. 

2.  When  the  sum  of  any  column  is  greater  than  9,  the  process  of  adding  the 
tens  to  the  next  column  is  called  currying. 

66.  PROOF.    There  are  two  principal  methods  of  proving 
Addition. 

1st.  By  varying  the  combinations. 

Begin  with  the  right  hand  or  unit  column,  and  add  the  figures 
in  each  column  in  an  opposite  direction  from  that  in  which  they 
were  first  added ;  if  the  two  results  agree,  the  work  is  supposed 
to  be  right. 

2d.  By  excess  of  9's. 

67.  This  method  depends  upon  the  following  properties  of  the 
number  9  :  * 

I.  If  a  number  be  divided  by  9,  the  remainder  will  be  the  same 
as  when  the  sum  of  its  digits  is  divided  by  9.     Therefore, 

II.  If  several  numbers  be  added,  the  excess  of  9's  in  the  sum 
must  be  equal  to  the  excess  of  9's  in  the  sum  of  all  the  digits  in 
the  numbers. 

1.  Add  34852,  24784,  and  72456,  and  prove  the  work  by  the 
excess  of  9's. 

OPERATION. 

34852 
24784 
72456  ...  8.  excess  of  9's  in  all  the  digits  of  the  numbers. 


132092  ...  8,      "  "       "    sum 

ANALYSIS.  Commencing  with  the  first  number,  at  the  left  hand,  we 
say  3  and  4  are  7,  and  8  are  15  ;  dropping  9,  the  excess  is  6,  which 
added  to  5,  the  next  digit,  makes  11;  dropping  9,  the  excess  is  2; 
then  2  and  2  are  4,  and  2  (the  left  hand  digit  of  the  second  number) 

*  For  a  demonstration  of  these  properties,  see  138,  IX. 


ADDITION  25 

are  6,  and  4  are  10 ;  dropping  9,  the  excess  is  1.  Proceeding  in  like 
manner  through  all  the  digits,  the  final  excess  is  8  ;  and  as  8  is  also 
the  excess  of  9's  in  the  sum,  the  work  of  addition  is  correct.  It  is 
evident  that  the  same  result  will  be  obtained  by  adding  the  digits  in 
columns  as  in  rows.  Hence,  to  prove  Addition  by  excess  of  9's :  — 
Commencing  at  any  figure,  add  the  digits  of  the  given  numbers 
in  any  order,  dropping  9  as  often  as  the  amount  exceeds  9.  If 
the  final  excess  be  equal  to  the  excess  of  9's  in  the  sum,  the  work 
is  right. 

NOTE. — This  method  of  proving  addition  by  the  excess  of  9's,  fails  in  the  fol- 
lowing cases:  1st,  when  the  figures  of  the  answer  are  misplaced;  2d,  when  the 
value  of  one  figure  is  as  much  too  great  as  that  of  another  is  too  small, 

EXAMPLES    FOR   PRACTICE 


(1.) 

(20 

(3.) 

(4.) 

8635 

1234567 

67 

24603 

2194 

723456 

123 

298765 

7421 

34565 

4567 

47321 

5063 

45666 

89093 

58653 

2196 

333 

654321 

5376 

1245 

90 

1234567 

340 

26754 

2038677 

5.  123+456+ 785+ 12-f  345+  901+ 567=how  many? 

6.  12345+67890+ 8763 +347 +  1037 +198760= how  many? 

7.  172+4005+3761  +  20472+367012+19762=how  many? 

8.  What  is  Jie  sum  of  thirty-seven  thousand  six,  four  hundred 
twenty-nine  thousand  nine,  and  two  millions  thirty-six  ? 

Ans.  2,466,051. 

9.  Add  eight  hundred  fifty-six  thousand  nine  hundred  thirty- 
three,  one  million  nine  hundred  seventy-six  thousand  eight  hun- 
dred fifty-nine,  two  hundred  three  millions  eight  hundred  ninety- 
five  thousand  seven  hundred  fifty-two.  Ans.  206,729,544. 

10.  What  is  the  sum  of   one  hundred   sixty-seven  thousand, 
three  hundred  sixty-seven  thoasand,  nine  hundred  six  thousand, 
two  hundred  forty-seven  thousand,  seventeen  thousand,  one  hun- 
dred six  thousand  three  hundred,  forty  thousand  forty-nine,  ten 
thousand  four  hundred  one?  Ans.  1,860,750. 

11.  What  number  of  square   miles   in  New  England,   there 
3 


26  SIMPLE  NUMBERS. 

being  in  Maine  31766,  in  New  Hampshire  9280,  in  Vermont 
10212,  in  Massachusetts  7800,  in  Rhode  Island  1306,  and  in 
Connecticut  4674?  Ans.  65,038. 

12.  The  estimated  population  of  the  above  States,  in  1855,  was 
as  follows;  Maine  653000,  New  Hampshire  338000,  Vermont 
327000,  Massachusetts  1133123,  Rhode  Island  166500,  and  Con* 
necticut  384000.  What  was  the  entire  population  ? 

13  At  the  commencement  of  the  year  1858  there  were  in  ope- 
ration in  the  New  England  States,  3751  miles  of  railroad;  in 
New  York,  2590  miles ;  in  Pennsylvania,  2546 ;  in  Ohio,  2946 ; 
in  Virginia,  1233 ;  in  Illinois,  2678  ;  and  in  Georgia,  1233.  What 
was  the  aggregate  number  of  miles  in  operation  in  all  these  States  ? 

14.  The  Grand  Trunk  Railway  is  962   miles  long,  and   cost 
$60000000 ;  the  Great  Western  Railway  is  229  miles  long,  and 
cost  $14000000;   the  Ontario,  Simcoe  and   Huron,  is  95  miles 
long,  and  cost  $3300000 ;  the  Toronto  and  Hamilton  is  38  miles 
long,  and  cost    $2000000.     What  is  the  aggregate  length,  and 
what  the  cost,  of  these  four  roads  ? 

Ans.  Length,  1,324  miles ;  cost,  $79,300,000, 

15.  A  man  bequeathed  his  estate  as  follows;   to  each  of  his 
two  sons,  $12450;    to  each  of  his  three  daughters,  $6500;    to 
his  wife,  $650  more  than  to  both  the  sons,  and   the  remainder, 
which  was  $1000  more  than  he  had  left  to  all  his  family,  he  gave 
to  benevolent  institutions.     What  was  the  whole  amount  of  his 
property?  Ans.  $140,900. 

16.  How  many  miles  from  the  southern  extremity  of  Lake 
Michigan  to  the   Gulf  of  St.  Lawrence,  passing  through  Lake 
Michigan,  330  miles;  Lake  Huron,  260  miles;  River  St.  C lair,  24 
miles;   Lake  St.  Clair,  20  miles;   Detroit  River,  23  miles;   Lake 
Erie,  260  miles;   Niagara  River,  34  miles;  Lake  Ontario,  180 
miles ;  and  the  River  St.  Lawrence,  750  miles  ? 

17.  The  United  States  exported  molasses,  in  the  year  1856,  to 
the  value  of  $154630;  in  1857,  $108003;  in  1858,  $115893; 
and  tobacco,  during  the  same  years  respectively,  to  the  value  of 
$1829207,  $1458553,  and  $2410224.     What  was  the  entire  value 
of  the  molasses  and  tobacco  exported  in  these  three  years  ? 


ADDITION.  27 

18.  The  population  of  Boston,  in  1855,  was  162629;   Provi- 
dence, 50000;  New  York,  629810;  Philadelphia,  408815;  Brook- 
lyn, 127618 ;  Cleveland,  43740 ;  and  New  Haven,  25000.     What 
was  the  entire  population  of  these  cities  ?  Ans.  1,447,612. 

19.  Iron  was  discovered  in  Greece  by  the  burning  of  Mount 
Ida,  B.  C.  1406 ;  and  the  electro-magnetic  telegraph  was  invented 
by  Morse,  A.  D.  1832.     What  period  of  time  elapsed  between  the 
two  events  ?  Ans.  3,238  years. 

20.  The  number  of  pieces  of  silver  coin  made  at  the  United 
States  Mint  at  Philadelphia,  in  the  year  1858,  were  as  follows : 
4628000  half  dollars,  10600000  quarter  dollars,  690000  dimes, 
4000000  half  dimes,  and  1266000  three-cent  pieces.     What  wag 
the  total  number  of  pieces  coined  ? 

21.  The  cigars  imported  by  the  United  States,  in  the  year  1856, 
were  valued  at  $3741460;  in  1857,  at  $4221096;  and  in  1858,  at 
$4123208.     What  was  the  total  value  of  the  importations  for  the 
three  years  ?  Ans.  $12,085,764. 

22.  In  the  appropriations  made  by  Congress  for  the  year  ending 
June  30,  1860,  were  the  following;   for  salary  and  mileage  of 
members  of  Congress,  $1557861 ;   to  officers  and  clerks  of  both 
Houses,  $1 57639 ;  for  paper  and  printing  of  both  Houses,  $170000 ; 
to  the  President  of  the  United  States,  $31450 ;  and  to  the  Vice 
President,  $8000.     WThat  is  the  total  of  these  items  ? 


ADDING   TWO   OR    MORE   COLUMNS  AT   ONE    OPERATION. 

68.   1.  What  is  the  sum  of  4632,  2553,  4735,  and  2863  ? 

OPERATION.          ANALYSIS.     Beginning  with  the  units  and  tens  of 

4632          the  number  last  written,  we  add  first  the  tens  above, 

2553          then  the  units,  thus ;  63  and  30  are  93,  and  5  are  98, 

4735          and  50  are  148,  and  3  are  151,  and  30  are  181,  and 

2  are  183.     Of  this  sum,  we  write  the  83  under  the 

14783          columns  added,  and  carry  the  1  to  the  next  columns, 

thus ;  28  and  1  are  29,  and  40  are  69,  and  7  are  76, 

and  20  are  96,  and  5  are  101,  and  40  are  141,  and  6  are  147,  whicb 

we  write  in  its  place,  and  we  have  the  whole  amount,  14783. 


28 


SIMPLE  NUMBERS. 


EXAMPLES    FOR   PRACTICE. 


(I-) 

8450 
5425 
8595 
6731 
7963 
5143 
4561 
6783 
4746 
2373 
3021 
7273 

71064 


(20 
75634 
86213 
92045 
73461 
34719 
26054 
19732 
84160 
97013 
34567 
43651 
52170 

719419 


(3-) 
123456 

47021 

82176 

570914 

379623 

7542 

25320 

57644 
908176 

73409 
3147 

67039 

2345467 


(40 

7349042 

2821986 

1621873 

236719 

401963 

67254 

45067 

910732 

6328419 

1437651 

9716420 

3191232 


34128358 


5.  What  is  the  total  number  of  churches,  the  number  of  persons 
accommodated,  and  the  value  of  church  property  in  the  United 
States,  as  shown  by  the  following  statistics  ? 

No.  of  No.  of  persons  Value  of 

churches.  accommodated.  church  property 

Methodist 12484  4220293  $14636671 

Baptist 8798  3134438  10931382 

Presbyterian 4591  2045516  14469889 

Congregational 1675  795677  7973962 

Episcopal 1430  631613  11261970 

Roman  Catholic 1269  705983  8973838 

Lutheran 1205  532100  2867886 

Christians 812  296050  845810 

Friends 715  283023  1709867 

Union 619  213552  690065 

Universalist 494  205462  1767015 

Free  Church 361  108605  252255 

Moravian 331  112185  443347 

German  Reformed 327  156932  965880 

Dutch  Reformed 324  181986  4096730 

Unitarian 244  137867  3268122 

Mennonite 110  29900  94245 

Tunkers 52  35075  46025 

Jewish.., 31  16575  371600 

Swedenborgian 15  5070  108100 


ADDITION. 


29 


6.  Give  the  amounts  of  the  productions  of  the  United  States  and 
Territories  for  the  year  1850,  as  expressed  in  the  following  columns : 


Alabama  
Arkansas  .  « 

Pounds  of 
Butter. 

4,008,811 
1,854,239 

Pounds  of 
Cheese. 

31,412 

30,088 

Pounds  of 
Wool. 

657,118 
182,595 

Bushels  of 
Wheat. 

294,044 
199,639 

California  
Columb.jDist. 
Connecticut... 
Delaware  •  •  • 

705 
14,872 
6,498,119 
1,055,308 

150 
1,500 
5,363,277 
3,187 

5,520 
525 
497,454 

57,768 

17,228 
17,370 
41,762 
482,511 

Florida  . 

371,498 

18,015 

23,247 

1,027 

Georgia  .  . 

4,640,559 

46,976 

990,019 

1,088,534 

Illinois  

12,526,543 

1,278,225 

2,150,113 

9,414,575 

Indiana  

12,881,535 

624,564 

2,610,287 

6,214,458 

Iowa  

2,171,188 

209,840 

•   373,898 

1,530,581 

Kentucky  .... 
Louisiana  

9  947,523 
'683,069 
9,243,811 

213,954 
1,957 
2,434,454 

2,297,433 
109,897 
1,364,034 

2,142,822 
417 

296,259 

Maryland  
Massachusetts 
Michigan  
Mississippi  .  .  . 
Missouri  

3,086,160 
8,071,370 
7,065,878 
4,346,234 
7,834,359 

3,975 

7,088,142 
1,011,492 
21,191 
203,572 

477,438 
585,138 
2,043,283 
559,619 
1,627,164 

4,494,680 
31,211 

4,925,889 
137,990 
2,981,652 

N.  Hampshire 
New  Jersey... 
New  York.... 
N.Carolina... 
Ohio  

6,977,056 
9,487,210 
79,766,094 
4,146,290 
34,449,379 

3,196,563 
365,756 
49,741,413 
95,921 
20,819,542 

1,108,476 
375,396 
10,071,301 
970,738 
10,196,371 

185,658 
1,601,190 
13,121,498 
2,130,102 
14,487,351 

Pennsylvania 
Rhode  Island 
S.  Carolina... 
Tennessee  .... 
Texas  

39,878,418 
995,670 
2,981,850 
8,139,585 
2,344,900 

2,505,034 
316,508 
4,970 
177,681 
95,299 

4,481,570 
129,692 
487,233 
1,364,378 
131,917 

15,367,691 
49 
1,066,277 
1,619,386 
41,729 

Vermont  
Virginia  
Wisconsin  .... 
Territories  .  .  . 

12,137,980 
11,089,359 
3,633,750 
295,984 

8,720,834 
436,292 
400,283 
73,826 

3,400,717 
2,860,765 
253,963 
71,894- 

535,955 
11,212,616 
4,286,131 
517,562 

9* 


30  SIMPLE    NUMBERS. 


SUBTRACTION. 

GO.  Subtraction  is  the  process  of  determining  the  difference 
between  two  numbers  of  tlte  same  unit  value. 

70.  The  Minuend  is  the  number  to  be  subtracted  from. 

71.  The  Subtrahend  is  the  number  to  be  subtracted. 

72.  The  Difference  or  Remainder  is  the  result  obtained  by 
the  process  of  subtraction. 

73.  When  the  given  numbers  contain  more  than  one  figure 
each,  the  method  of  subtraction  depends  upon  the  following  prin- 
ciples : 

I.  If  the  units  of  each  order  in  the  subtrahend  be  taken  sepa- 
rately from  the  units  of  like  order  in  the  minuend,  the  sum  of  the 
differences  must  be  equal  to  the  entire  difference  of  the  given 
numbers.     (Ax.  10.) 

II.  If  both  minuend  and  subtrahend  be  equally  increased,  the 
remainder  will  not  be  changed. 

1.    From  928  take  275. 

OPERATION.  ANALYSIS.  We  first  subtract  5  units  from 
Minuend,  928  8  units,  and  obtain  3  units  for  a  partial  re- 
Subtrahend;  275  mainder.  As  we  cannot  take  7  tens  from  2 

Remainder  653  tenS'  WG   a<^  ^  tenS  *°  *^6  ^  ten8'  makmg 

12  tens;   then  7  tens  from  12  tens  leave  5 

tens,  the  second  partial  remainder.  Now,  since  we  added  10  tens, 
or  1  hundred,  to  the  minuend,  we  will  add  1  hundred  to  the  subtra- 
hend, and  tne  true  remainder  will  not  be  changed  (II) ;  thus,  1 
hundred  addsd  to  2  hundreds  makes  3  hundreds,  and  this  sum  sub- 
tracted from  9  hundreds  leaves  6  hundreds ;  and  we  have  for  the  total 
remainder,  653. 

NOTK. — The  process  of  adding  10  to  the  minuend  is  sometimes  called  bor- 
rowing 10,  and  that  of  adding  1  to  the  next  figure  of  the  subtrahend,  carrying  1. 

74.  From  these  principles  and  illustrations  we  deduce  the 
following 

RULE.     I.   Write  the  less  number  under  the  greater,  placing  units 
vf  the  same  order  under  each  other 


SUBTRACTION.  31 

II.  Begin  at  the  right  hand^and  take  each  figure  of  the  subtra- 
hend from  the  figure  above  it,  and  write  the  result  underneath. 

III.  If  any  figure  in  the  subtrahend  be  greater  than  the  corres- 
ponding figure  above  it,  add  10  to  that  upper  figure  before  sub- 
tracting,  and  then  add  1  to  the  next  left-hand  figure  of  the  subtra- 
hend. 

7«5,  PROOF.  It  is  evident  that  the  subtrahend  and  remainder 
must  together  contain  as  many  units  as  the  minuend ;  hence,  to 
prove  subtraction,  we  have  three  methods  : 

1st.  Add  the  remainder  to  the  subtrahend;  the  sum  will  be 
equal  to  the  minuend.  Or, 

2d.  Subtract  the  remainder  from  the  minuend ;  the  difference 
will  be  equal  to  the  subtrahend.  Or, 

3d.  Find  the  excess  of  9's  in  the  remainder  and  subtrahend 
together,  and  it  will  be  equal  to  the  excess  of  9's  in  the  minuend. 

EXAMPLES   FOR   PRACTICE. 

(1.)      (2.)       (3.)       (4.) 
From    47965    103767    57610218    89764321 
Take    26714    98731     8306429    83720595 

Hem.    21251     5036    49303789    6043726 

5.  From  180037561  take  5703746. 

6.  From  2460371219  take  98720342. 

7.  89037426175  —  2435036749  =  how  many? 

8.  10000033421—999044110  =  how  many? 

9.  A  certain  city  contains  146758  inhabitants,  which  is  3976 
more  than  it  contained  last  year;    how  many  did  it  contain  last 
year?  Ans.  142,782. 

10.  The  first  newspaper  published  in  America  was  issued  at 
Boston  in  1704;  how  long  was  that  before  the  death  of  Benjamin 
Franklin,  which  occurred  in  1790  ? 

11.  A  merchant  sold  a  quantity  of  goods  for   $42017,  which 
was  $1675  more  than  they  cost  him;  how  much  aid  they  cost 
him?  Ans.  $40,342. 

12.  In  1858  the  exports  of  the  United  States  amounted  to 


32  SIMPLE  NUMBERS. 

$324644421,  and  the  imports  to  $282613150;  how' much  did  the 
exports  exceed  the  imports  ?  Ans.  $42,031,271. 

13.  In  1858  the  gold  coinage  of  the  United  States  amounted  to 
$52889800,  and  the  silver  to  $8233287;  how  much  did  the  gold 
exceed  the  silver  coinage  ? 

14.  The  South  in  1850  produced  978311690  pounds  of  cotton, 
valued  at  $101834616,  and  237133000  pounds  of  sugar  valued  at 
$16599310;  how  much  did  the  cotton  exceed  the  sugar  in  quan- 
tity and  in  value  ?         Ans.  741,178,690  pounds;  $85,235,306. 

15.  The  area  of  the  Chinese  Empire  is  5110000  square  miles, 
and  that  of  the  United  States  2988892  square  miles;  the  esti- 
mated population  of  the  former  is  340000000,  and  that  of  the 
latter  in  1850  was  23363714.     What  is  the  difference  in  area  and 
in  population? 

16.  The  population  of  London  in  1850  was  2362000,  and  that 
of  New  York  city  515547;  how  many  more  inhabitants  had  London 
than  New  York  ?  Ans.     1,846,453. 

17.  The  total  length  of  railroads  in  operation  in  the  United 
States,  January  1,  1859,  was  27857  miles,  and  the  total  length  of 
the  canals  5131  miles;  how  many  miles  more  of  railroad  than  of 
canal?  Ans.  22,726, 

18.  The  entire  deposit  of  domestic  gold  at  the  United  States 
Mint  and  its  branches,  to  June,  1859,  was  $470341478,  of  which 
$451310840  was  from  California;  how  much  was  received  from 
other  sources  ?  Ans.  $19,030,638. 

19.  During  the  year  ending  September  30,  1858,  the  number 
of  letters  exchanged  between  the  United  States  and  Great  Britain 
were  1765015  received,  and  1603609  sent;  between  the  United 
States  and  France,  624795  received,  and  639906  sent.    How  many 
letters  did  the  exchange  with  Great  Britain  exceed  those  with 
France?  Ans.  2,103,923. 

20.  The  Southern  States  in  1850  had  a  population  of  66960(51, 
the   Middle  States  6624988,  and  the  Eastern  States  2728116; 
how  many  more  inhabitants  had  the  Middle  and  Eastern  States 
than  the  Southern  States  ? 

21.  Having  $20000,  I  wish  to  know  how  much  more  I  must 


SUBTRACTION.  33 

accumulate  to  be  able  to  purchase  a  piece  of  property  worth  $23470, 
aDd  have  $5400  left?  Ans.  $8,870. 

22.  A  has  $3540  more  than  B,  and  $1200  less  than  C,  who  has 
820600 ;  D  has  as  much  as  A  and  B  together.  How  much  has  D  ? 

Ans.  $35/260. 

TWO    OR    MORE    SUBTRAHENDS. 

76.  Two  or  more  numbers  may  be  taken  from  another  a!  a 
single  operation,  as  shown  by  the  following  example : 

I.  A  man  having  1278  barrels  of  flour,  sold  236  barrels  to  A, 
362  to  B,  and  387  to  C ;  how  many  had  he  left  ? 

OPERATION.  ANALYSIS.     Since  the  remainder  sought, 

Minuend,  1278          added  to  the  subtrahends,  must  be  equal  to 

9  the  minuend,  we  add  the  columns  of  the 

\     o/?«  subtrahends,  and  supply  such  figures  in  the 

Subtrahends,    J      doJ  .     .  .  .       ,        .  . 

l  ooy  remainder  as,  combined  with  these  sums, 
will  produce  the  minuend.  Thus,  7  and  2 
are  9,  and  6  are  15,  and  3  (supplied  in  the 
remainder  sought)  are  18 ;  then,  carrying 

the  tens'  figure  of  the  18,  1  and  8  are  9,  and  6  are  15,  and  3  are  18, 
and  9  (supplied  in  the  remainder)  are  27;  lastly,  2  to  carry  to  3  are 
5,  and  3  are  8,  and  2  are  10,  and  2  (supplied  in  the  remainder)  are 
12 ;  and  the  whole  remainder  is  293.  Hence,  the  following 

RULE.  T.  Having  written  the  several  subtrahends  under  the  min- 
uend, add  the  first  column  of  the  subtrahends,  and  supply  such  a 
figure  in  the  remainder  sought,  as,  added,  to  this  partial  sum,  will 
give  an  amount  having  for  its  unit  figure  the  figure  above  in  the 
minuend. 

II.  Carry  the  tens'1  figure  of  this  amount  to  the  next  column  of 
the  subtrahends,  and  proceed  as  before  till  the  entire  remainder  is 
obtained. 

EXAMPLES    FOR   PRACTICE. 

(!•) 
From 

Take 
Hem. 


(1.) 

47962 

(20 
127368 

(3.) 
903486 

(40 
2503734 

21435 
15672 
456 

56304 

4782 
9156 

430164 
132875 
67321 

89763 
94207 
237564 

10399 

57126 

273126 

2082200 

34  SIMPLE  NUMBERS. 

5.  From  4568  take  1320  -f  275  +  320. 

6.  Subtract  1200  +  750  -f  96  from  4756  -f  575  -f  140  +  84. 

7.  A  man  bought  four  city  lots,  for  which  he  paid  $15760.     Foi 
the  first  he  paid  $2175,  for  the  second  $3794,  and  for  the  third 
$4587;  how  much  did  he  pay  for  the  fourth  ?       Ans.  $5,204. 

8.  John  Wise  owns  property  to  the  amount  of  $75860;  of  which 
he  has  $45640  invested  in  real  estate,  $25175  in  personal  property, 
and  the  remainder  he  has  in  bank;  how  much  has  he  in  bank  ? 

9.  Lake  Huron  contains   20000  square  miles;   by  how  much 
does  it  exceed  the  area  of  Lake  Erie  and  Lake  Ontario,  the  former 
containing  11000  square  miles,  and  the  latter  7000  ? 

Ans.  2000  square  miles. 

10.  In  the  year  1852,  there  arrived  in  the  United  States  398470 
immigrants,  of  whom  157548  were  born  in  Ireland,  and  143429 
were  born  in  Germany;  how  many  were  born  in  other  countries? 

Ans.  97,493. 

11.  The  entire  amount  of  coinage  in  the  United  States  for  the 
year  ending  June,  1858,  was  $61357088,  of  which  $52889800  was 
of  gold,  $234000  of  copper,  and  the  remainder  of  silver;  how  much 
was  of  silver? 

12  A  speculator  gained  $5760,  and  afterward  lost  $2746 ;  at 
another  time  he  gained  $3575,  and  then  lost  $4632.  How  much 
did  his  gains  exceed  his  losses?  Ans.  $1,957. 

13.  The  Eastern  States  have  an  area  of  65038  square  miles, 
the  Middle  States  114624  square  miles,  and  the  Southern  States 
643166  square  miles;   how  many  more  square  miles  have  the 
Southern  than  the  Middle  and  Eastern  States  ? 

14.  The  entire  revenue  of  the  United  States  Post  Office  Depart- 
ment for  the  year  ending  Sept.  30,  1858,  was  $8186793,  of  which 
sum  $5700314  was  received  for  stamps  and  stamped  letters,  and 
$904299  for  letter-postage  in  money;  how  much  was  received  from 
all  other  sources  ?  Ans.  $1,582,180. 

15.  The  total  expenditures  oi  the  Department  for  the  same  year 
were  $12722470,  of  which  sum  $7821556  was  paid  for  the  trans- 
portation of  inland  mails,  $424497  for  the  transportation  of  foreign 
mails  and  $2355016  as  compensation  to  postmasters;  how  much 
was  expended  for  all  other  purposes?  $2,121,401. 


MULTIPLICATION.  35 


MULTIPLICATION. 

7  To  Multiplication  is  the  process  of  taking  one  of  two  given 
numbers  as  many  times  as  there  are  units  in  the  other, 

78t    The  Multiplicand  is  the  number  to  be  taken. 

79.  The  Multiplier  is  the  number  which  shows  how  many 
times  the  multiplicand  is  to  be  taken. 

8O  The  Product  is  the  result  obtained  by  the  process  of  mul- 
tiplication. 

81.  The  Factors  are  the  multiplicand  and  multiplier. 

NOTES.  —  1.  Factors  are  producers,  and  the  multiplicand  and  multiplier  are 
called  factors  because  they  produce  the  product. 

2.  Multiplication  is  a  short  method  of  performing  addition  when  the  numbers 
to  be  added  are  equal. 


The  method  of  multiplying  when  either  factor  contains 
more  than  one  figure,  depends  upon  the  following  principles  : 

It  is  evident  that  5  units  taken  3  times  is  the  same  as  3  units 
taken  5  times  ;  and  the  same  is  true  of  any  two  factors      Hence, 

I.  The  product  of  any  two  factors  is  the  same,  whichever  is  used 
as  the  multiplier.     If  units  be  multiplied  by  units,  the  product 
will  be  units  ;  if  tens  be  multiplied  by  units,  or  units  by  tens,  the 
product  will  be  tens;  and  so  on.     That  is, 

II.  If  either  factor  be  units  of  the  first  order,  the  product  will 
be  units  of  the  same  order  as  the  other  factor. 

III.  If  the  units  of  each  order  in  the  multiplicand  be  taken  sepa- 
rately as  many  times  as  there  are  units  in  the  multiplier,  the  sum 
of  the  products  must  be  equal  to  the  entire  product  of  the  given 
numbers,     (Ax.  10). 

1.   Multiply  346  by  8. 

OPERATION.  ANALYSIS.     In  this   example  it  is  re- 

Multiplicand,          346          quired  to  take  346  eight  times.     If  we  take 
Multiplier,  8          the  units  of  each  order  8  times,  we  shall 


Product  2768          ta^e  *^e   en^re  number   8   times,  (III). 

Therefore,  commencing  at  the  right  hand. 

we  say;  8  times  6  units  are  48  units,  or  4  tens  and  8  units  ;  writing 
the  8  units  in  the  product  in  units'  place,  we  reserve  the  4  tens  to 
add  to  the  next  product  ;  8  times  4  tens  are  32  tens,  and  the  4  tens 


36  SIMPLE  NUMBERS. 

reserved  in  the  last  product  added,  are  36  tens,  or  3  hundreds  and  6 
tens ;  we  write  the  6  tens  in  the  product  in  tens'  place,  and  reserve 
the  3  hundreds  to  add  to  the  next  product ;  8  times  3  hundreds  are 
24  hundreds,  and  the  3  hundreds  reserved  in  the  last  product  added, 
are  27  hundreds,  which  being  written  in  the  product,  each  figure  in 
the  plaoe  of  its  order,  gives,  for  the  entire  product,  2768. 

2.   Multiply  758  by  346. 

OPERATION.  ANALYSIS.     In  this  example  the  multiplicand  is 

758  to  be  taken  346  times,  which  may  be  done  by 

346  taking  the  multiplicand  separately  as  many  times 

JTTo  as  there  are  units  expressed  by  each  figure  of  the 

3032  multiplier.     758   multiplied   by  6  units   is  4548 

2274  units,  (II) ;  758  multiplied  by  4  tens  is  3032  tens, 

262268  ^II)'  which  we  write  with  its  lowest  order  in  tens' 

place,  or  under  the  figure  used  as  a  multiplier; 

758  multiplied  by  3  hundreds  is  2274  hundreds,  (II),  which  we  write 

with  its  lowest  order  in  hundreds7  place.     Since  the  sum  of  these 

products  must  be  the  entire  product  of  the  given  numbers,  (III),  we 

add  the  results,  and  obtain  262268,  the  answer. 

NOTES. — 1.  When  the  multiplier  contains  two  or  more  figures,  the  several  re- 
sults obtained  by  multiplying  by  each  figure  are  called  partial  products. 

2.  When  there  are  ciphers  between  the  significant  figures  of  the  multiplier, 
Dass  over  them,  and  multiply  by  the  significant  figures  only. 

83.  From  these  principles  and  illustrations  we  deduce  the  fol- 
fowing  general 

RULE.  I.  Write  the  multiplier  under  the  multiplicand,  placing 
units  of  the  same  order  under  each  other. 

II.  Multiply  the  multiplicand  by  each  figure  of  the  multiplier 
successively,  beginning  with  the  unit  figure,  and  write  the  first  figure 
of  each  partial  product  under  the  figure  of  the  multiplier  used, 
writing  down  and  carrying  as  in  addition. 

III.  If  there  are  partial  products,  add  them,  and  their  sum  will 
be  the  product  required . 

NOTE.  — The  multiplier  denotes  simply  the  number  of  times  the  multiplicand 
is  to  be  taken ;  hence,  in  the  analysis  of  .1  problem,  the  multiplier  must  be  con- 
sidered as  abstract,  though  the  multiplicand  may  be  either  abstract  or  concrete. 

84.  PROOF.   There  are  two  principal  methods  of  proving 
multiplication 


MULTIPLICATION.  37 

1st.  By  varying  the  partial  products. 

Invert  the  order  of  the  factors  ;  that  is,  multiply  the  multiplier 
by  the  multiplicand  :  if  the  product  is  the  same  as  the  first  result, 
the  work  is  correct. 

2d.  By  excess  of  9's. 

85.  The  illustration  of  this  method  depends  upon  the  following 
principles  : 

I.  If  the  excess  of  9's  be  subtracted  from  a  number,  the  re- 
mainder will  be  a  number  having  no  excess  of  9's. 

II.  If  a  number  having  no  excess  of  9's  be  multiplied  by  any 
number,  the  product  will  have  no  excess  of  9's. 

1.  Let  it  t>e  required  to  multiply  473  by  138. 

OPERATION.  ANALYSIS.     The   excess   of 

473  =  468  -f  5  9's  in  473  is  5,  and  473  =  468 

138  =135-f3  +5,  of  which  the  first  part, 


,  AQ         QP  _  '  no  excess  of  9's, 


Partial  5x135=        675  W-        The    eXC688    °f    9'S     in 

products.    1  468  X  3      =    1404        138  is  3,  and  138  =  135  -f  3,  of 
[      5x3      =         15         which  the  first  part,  135,  con- 

Entire  product,  65274        tains^  no   excess    of   9's,    (I). 

Multiplying  both  parts  of  the 

multiplicand  by  each  part  of  the  multiplier,  we  have  four  partial  pro^ 
ducts,  of  which  the  first  three  have  no  excess  of  9's,  because  each  con- 
tains a  factor  having  no  excess  of  9's,  (II).  Therefore,  the  excess 
of  9's  in  the  entire  product  must  be  the  same  as  the  excess  of  9's  in 
the  last  partial  product,  15,  which  we  find  to  be  1  +  5  =  6.  The  same 
may  be  shown  of  any  two  numbers.  Hence,  to  prove  multiplication 
by  excess  of  9's, 

Find  the  excess  of  9's  in  each  of  the  two  factors,  and  multiply 
them  together;  if  the  excess  of  9's  in  this  product  is  equal  to  the 
excess  of  9's  in  the  product  of  the  factors,  the  work  is  supposed 
to  be  right. 

NOTE.  —  If  the  excess  of  9's  in  either  factor  is  0,  the  excess  of  9's  in  the  pro- 
duct will  be  0,  (II). 

EXAMPLES    FOR   PRACTICE. 

(1.)  (2.)                  (3.)  (4.) 

Multiply     475  3172                9827  7198 

By                  9  14  _  84  216 

Prod.        4275  44408  825468  1554768 


38  SIMPLE  NUMBERS. 

i 

5  Multiply  31416  by  175.  Ans.  5497800 

6.  Multiply  40930  by  779.  Ans.  31884470. 

7.  Multiply  46481  by  936. 

8.  Multiply  15607  by  3094. 

9.  Multiply  281216  by  978.  Ans.  275029248 

10.  Multiply  30204  by  4267.  Ans.  128,880,468. 

11.  What  is  the  product  of  4444  x  2341  ? 

Ans.  10,403,404. 

12.  What  is  the  product  of  4567  X  9009  ? 

Ans.  41,144,103. 

13.  What  is  the  product  of  2778588  x  9867? 

Ans.  27,416,327,796. 

14.  What  is  the  product  of  7060504  x  30204  ? 

Ans.  213,255,462,816. 

15.  What  will  be  the  cost  of  building  276  miles  of  railroad  at 
$61320  per  mile  ?  Ans.  $16,924,320. 

16.  If  it  require  125  tons  of  iron  rail  for  one  mile  of  railroad, 
how  many  tons  will  be  required  for  196  miles  ? 

17.  A  merchant  tailor  bought  36   pieces  of  broadcloth,  each 
piece  containing  47  yards,  at  7  dollars  a  yard ;  how  much  did  he 
pay  for  the  whole  ?  Ans.  $11,844. 

18.  The  railroads  in  the  State  of  New  York,  in  operation  in 
1858,  amounted  to  2590  miles  in  length,  and  their  average  cost 
was  about  $52916  per  mile;  what  was  the  total  cost  of  the  rail- 
roads in  New  York '(  Ans.  $137,052,440. 

19.  The  Illinois  Central  Railroad  is  700  miles  long,  and  cost 
$45210  per  mile ;  what  was  its  total  cost  ? 

20.  The  salary  of  a  member  of  Congress  is  $3000,  and  in  1860 
there  were  303  members ;  how  much  did  they  all  receive  ? 

21.  The  United  States  contain  an  area  of  2988892  square  miles, 
and  in  1850  they  contained  8  inhabitants  to  each  square  mile; 
what  was  their  entire  population  ?  Ans.   23,911,136. 

22  Great  Britain  and  Ireland  have  an  area  of  118949  square 
miles,  and  in  1850  they  contained  a  population  of  232  ^o  the  square 
mile;  what  was  their  entire  population?  Ans.  27,596,168. 

23.  The  national  debt  of  France  amounts  to  $32  for  each  indi- 


MULTIPLICATION.  39 

vidual,  and  the  population  in  1850  was  35781628 ;  what  was  the 
entire  debt  of  France  ?  Ans.   1,145,012,096. 

POWERS    OF   NUMBERS. 

8G.  We  have  learned  (15)  that  a  power  is  the  product  arising 
from  multiplying  a  number  by  itself,  or  repeating  it  any  number 
of  times  as  a  factor;  (16),  that  a  root  is  a  factor  repeated  to  pro- 
duce a  power;  and  (4O)  an  index  or  exponent  is  the  number  in- 
dicating the  power  to  which  a  number  is  to  be  raised. 

87.  The  First  Power  of  any  number  is  the  number  itself,  or 
the  root;  thus,  2,  3,  5,  are  first  powers  or  roots. 

88.  The  Second  Power,  or  Square,  of  a  number  is  the  pro- 
duct arising  from  using  the  number  two  times  as  a  factor;  thus, 
22^2x2  =  4;  52=5x5=:25. 

89.  The  Third  Power,  or  Cube,  of  a  number  is  the  product 
arising  from  using  the  number  three  times  as  a  factor;    thus, 
43=4x  4x4  =  64. 

00.  The  higher  powers  are  named  in  the  order  of  their  num- 
bers, as  Fourth  Power,  Fifth  Power,  Sixth  Power,  etc. 

01.  I-  What  is  the  third  power  or  cube  of  23  ? 

OPERATION.  ANALYSIS.     We  multiply  23 

23  X  23  x  23  =  12167  bJ  23'  and  the  product  by  23; 

and,  since  23  has  been  taken  3 

times  as  a  factor,  the  last  product,  12167,  must  be  the  third  power  or 
cube  of  23.     Hence, 

RULE.  Multiply  the  number  by  itself  as  'many  times,  less  1,  as 
there  are  units  in  the  exponent  of  the  required  power. 

NOTE. — The  process  of  producing  any  required  power  of  a  number  by  multi- 
plication is  called  Involution. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  square  of  72  ?  Ans    5184. 

2.  What  is  the  fifth  power  of  12  ?  Ans.  248832. 

3.  What  is  the  cube  of  25  ? 

4.  What  is  the  seventh  power  of  7  ?  Ans.  823543. 


40  SIMPLE  NUMBERS. 

5.  What  is  the  fourth  power  of  19  ?  Am.  130321. 

6.  Required  the  sixth  power  of  3.  Ans.  729. 

7.  Find  the  powers  indicated  in  the  following  expressions; 
96,  IP,  182,  1254,  7862,  946,  1004,  173,  251.' 

8.  Multiply  83  by  152.  Ans.  115200. 

9.  What  is  the  product  of  252  X  34? 

10.  7s  ^  200  =  4*  X  II2,  and  how  many  ?          Ans.  37,624. 

GENERAL  PRINCIPLES  OF  MULTIPICATION. 

93.  There  are  certain  general  principles  of  multiplication,  of 
use  in  various  contractions  and  applications  which  occur  in  sub- 
sequent portions  of  this  work.  These  relate,  1st,  to  changing  the 
factors  by  addition  or  subtraction;  2d,  to  the  use  of  successive 
factors  in  continued  multiplication. 

CHANGING    THE   FACTORS   BY   ADDITION    OR    SUBTRACTION. 

93.  The  product  is  equal  to  either  factor  taken  as  many  times 
as  there  are  units  in  the  other  factor.     (83,  I).     Hence, 

I.  Adding  1  to  either  factor,  adds  the  other  factor  to  the  pro- 
duct. 

11.  Subtracting  1  from  either  factor,  subtracts  the  other  factor 
from  the  product.     Hence, 

III.  ADDING  any  number  to  either  factor,  INCREASES  the  pro* 
duct  by  as  many  times  the  other  factor  as  there  are  units  in  tht 
number  added;  and  SUBTRACTING  any  number  from  either  factor, 
DIMINISHES  the  product  by  as  many  times  the  other  factor  as  there 
are  units  in  the  number  subtracted. 

CONTINUED  MULTIPLICATION. 

94.  A  Continued  Multiplication  is  the  process  of  finding  the 
product  of  three  or  more  factors,  by  multiplying  the  first  by  the 
second,  this  result  by  the  third,  and  so  on. 

95.  To   show   the    nature    of    continued   multiplication,  we 
observe : 

1st.  If  any  number,  as  17,  be  multiplied  by  any  other  number, 
as  3,  the  result  will  be  3  times  17;  if  this  result  be  multiplied  by 


MULTIPLICATION.  41 

another  number,  as  5,  the  new  product  will  be  5  times  3  times  17, 
which  is  evidently  15  times  17.  Hence,  17  X  3  X  5  =  17  X  15; 
the  same  reasoning  would  extend  to  three  or  more  multipliers. 

2d.  Since  5  times  3  is  equal  to  3  times  5,  (82, 1),  it  follows  that 
17  multiplied  by  5  times  3  is  the  same  as  17  multiplied  by  3  times 
5;  or  17  X  3  x  5  =  17  X  5  X  3.  Hence,  the  product  is  not 
changed  by  changing  the  orders  of  the  factors. 

These  principles  may  be  stated  as  follows : 

I.  If  a  given  number  be  multiplied  by  several  factors  in  con- 
tinued multiplication,  the  result  will  be  the  same  as  if  the  given 
number  were  multiplied  by  the  product  of  the  several  multipliers. 

II.  The  product  of  several  factors  in  continued  multiplication 
will  be  the  same,  in  whatever  order  the  factors  are  taken. 

CONTRACTIONS  IN  MULTIPLICATION. 
CASE  I. 

96.  When  the  multiplier  is  a  composite  number. 

A  Composite  Number  is  one  that  may  be  produced  by  multi- 
plying together  two  or  more  numbers.  Thus,  18  is  a  composite 
number,  since  (5  x  3  =  18 ;  or,  9  x  2  =  18 ;  or,  3  x  3  X  2  =  18. 

97.  The  Component  Factors  of  a  number  are  the  several 
numbers  which,  multiplied  together,  produce  the  given  number; 
thus,  the  component  factors  of  20  are  10  and  2  (10  X  2  =  20); 
or,  4  and  5  (4  x  5  =  20) ;  or,  2  and  2  and  5  (2  x  2  x  5  =  20). 

NOTE. — The  pupil  must  not  confound  the  factors  with  the  part*  of  a  number. 
Thus,  the  factors  of  which  12  is  composed,  are  4  and  3  (4  X  3  =  12) ;  while  the 
parts  of  which  12  is  composed  are  8  and  4  (8 +  4=  12);  or  10  and  2  (10  +  2  =  12). 
The  factors  are  multiplied,  while  the  parts  are  added,  to  produce  the  number. 

98.  1.    Multiply  327  by  35. 

ANALYSIS.  The  factors  of  35  are  7  and  5. 
"We  multiply  327  by  7,  and  this  result  by  5, 
and  obtain  11445,  which  must  be  the  same 
as  the  product  of  327  by  5  times  7,  or  35. 
(95,  I).  Hence  we  have  the  following 


42  SIMPLE  NUMBERS. 

RULE.  I.  Separate  the  composite  number  into  two  or  more 
factors. 

II.  Multiply  the  multiplicand  ~by  one  of  these  factors,  and  that 
product  by  another,  and  so  on  until  all  the  factors  have  been  used 
successively  ;  the  last  product  will  be  the  product  required. 

NOTE. — The  factors  may  be  used  in  any  order  that  is  most  convenient,  (95,  II). 
EXAMPLES    FOR   PRACTICE. 

1.  Multiply  736  by  24.  Ans.  17664. 

2.  Multiply  538  by  56.  Ans.  30128. 

3.  Multiply  27865  by  84. 

4.  Multiply  7856  by  144.  Ant.  1131264. 

5.  What  will  56  horses  cost  at  185  each  ?  » 

6.  If  a  river  discharge  17740872  cubic  feet  of  water  in  one 
hour,  how  much  will  it  discharge  in  96  hours  ? 

Ans    1703123712  cubic  feet. 

CASE   II. 

99.   When  the  multiplier  is  a  unit  of  any  order. 

If  we  annex  a  cipher  to  the  multiplicand,  each  figure  is  removed 
one  place  toward  the  left,  and  consequently  the  value  of  the  Whole 
number  is  increased  tenfold,  (57,  III).  If  two  ciphers  are 
annexed,  each  figure  is  removed  two  places  toward  the  left,  and 
the  value  of  the  number  is  increased  one  hundred  fold ;  and  every 
additional  cipher  increases  the  value  tenfold.  Hence,  the 

RULE  Annex  as  many  ciphers  to  the  multiplicand  as  there  are 
ciphers  in  the  multiplier. 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  364  by  100.  Ans.  36400. 

2.  Multiply  248  by  1000.  Ans.  248000. 

3.  What  cost  1000  head  of  cattle  at  50  dollars  each? 

4.  Multiply  one  million  by  one  hundred  thousand  ? 

5.  How  many  letters  will  there  be  on  100  sheets,  if  each  sheet 
have  100  lines,  and  each  line  100  letters  ?  Ans.  1000000. 


MULTIPLICATION.  43 

CASE   III. 

100.  When  there  are  ciphers  at  the  right  hand  of  one 
or  both  of  the  factors. 

1.  Multiply  7200  by  40. 

OPERATION.  ANALYSIS.     The   multiplicand,  factored,  is 

7200  equal  to  72  X  100;  the  multiplier,  factored,  is 

40  equal  to  4  X  10  ;  and  as  these  factors  taken  in 

288000  any  order  wiH  giye  tne  same  product,  (95,  II), 

we  first  multiply  72  by  4,  then  this  product 

by  100  by  annexing  two  ciphers,  and  this  product  by  10  by  annexing 
one  cipher.     Hence,  the  following 

RULE.  Multiply  the  significant  figures  of  the  multiplicand  by 
those  of  the  multiplier,  and  to  the  product  annex  as  many  ciphers 
as  there  are  ciphers  on  the  right  of  both  factors. 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  740  by  300.  Ans.  222000. 

2.  Multiply  36000  by  240.  Ans.  8640000. 

3.  Multiply  20700  by  500. 

4.  Multiply  4007000  by  3002.  Ans.  12029014000. 
5  Multiply  300200  by  640. 

CASE   IV. 

101.  "When  one  part  of  the  multiplier  is  a  factor  of 
another  part. 

1.  Multiply  4739  by  357. 

OPERATION.  ANALYSIS.     In  this  example,  7,  one 

4700  part  of  the  multiplier,  is  a  factor  of 

35,  the  other  part.     We  first  find,  in 

the  usual  manner,  the  product  of  the 

33173  Prod,  by  7  units.        muitiplicand  by  the  7  units;  multi- 

165866       Prod,  by  35  tens 


1691823   Ans.  the  first  figure  of  the  result  in  tens' 

place,  we  obtain  the  product  of  the 

multiplicand  by  7  X  5  X  10  =  35  tens  ;  and  the  sum  of  these  two  par- 
tial products  must  be  the  whole  product  required. 


44  SIMPLE  NUMBERS. 

2.  Multiply  58327  by  21318. 

OPERATION.  ANALYSIS.     In  this  exam 

58397  P^e'  ^e  ^  hundreds  is  a  factor 

2131$  of  18,  the  part  on  the  right  of 

it,  and  also  of  21,  the  part  on 

Prod,  by  3  hundreds.  ^  lefl  of  ^       We  fipgt          j_ 

1049886    Prod,  by  18  units.  ,.    , 

tiply  by  3,  writing  the  first 

1224867  Prod,  by  21  thousands.         . L    J       *       ' 

* !__  figure    in    hundreds'    place  ; 

1243414986  Ans.  multiplying  this  product  by 

6,  and  writing  the  first  figure 

in  units'  place,  we  obtain  the  product  of  the  multiplicand  by  3  X  6  = 
18  units ;  multiplying  the  first  partial  product  by  7,  and  writing  the 
first  figure  in  thousands'  place,  we  obtain  the  product  of  the  multipli- 
cand by  7  X  3  X  1000  =  21  thousands ,  and  the  sum  of  these  three 
partial  products  must  be  the  entire  product  required. 

NOTE. — The  product  obtained  by  multiplying  any  partial  product  is  called  a 
derived  product. 

1O2.    From  these  illustrations  we  have  the  following 

RULE.     I.  Find  the  product  of  the  multiplicand  by  some  figure 

of  the  multiplier  which  is  a  factor  of  one  or  more  parts  of  the 

multiplier. 

II.  Multiply  this  product  ~by  that  factor  which,  taken  with  the 
figure  of  the  multiplier  first  used,  will  produce  other  parts  of  the 
multiplier,  and  write  the  .first  figure  of  each  result  under  the  first 
figure  of  the  part  of  the  multiplier  thus  used. 

III.  In  like  manner,  find  the  product,  either  direct  or  derived, 
for  every  figure  or  part  of  the  multiplier;  the  sum  of  all  the  pro- 
ducts will  be  the  whole  product  required. 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  5784  by  246.  Ans.  1422864. 

2.  Multiply  3785  by  721.  Ans.  2728985. 

3.  Multiply  472856  by  54918.  Ans.  25968305808. 

4.  Multiply  43785  by  7153.  Ans    313194105. 

5.  Multiply  573042  by  24816.  Ans.  14220610272. 

6.  Multiply  78563721  by  127369. 

7.  Multiply  43  T25652  by  51879 14. 


MULTIPLICATION.  45 

8.  Multiply  3578426785  by  64532164. 

9.  Multiply  2703605  by  4249784. 

10.  What  is  the  product  of  9462108  multiplied  by  16824? 

Ans.  159,190,504,992. 

EXAMPLES   COMBINING   THE   PRECEDING    RULES. 

1.  A  man  bought  two  farms,  one  containing  175  acres  at  $28 
per  acre,  and  the  other  containing  320  acres  at  837   per  acre; 
what  was  the  cost  of  both  ?  Ans.  $16,740. 

2.  If  a  man  receive  $1200  salary,  and  pay  $364  for  board, 
$275  for  clothing,  $150  for  books,  and  $187  for  other  expenses, 
how  much  can  he  save  in  5  years?  Ans.  $1,120. 

3.  Two  persons  start  from  the  same  point,  and  travel  in  oppo- 
site directions;  one  travels  29  miles  a  day,  and  the  other  32  miles. 
How  far  apart  will  they  be  in  17  days?  Ans.  1,037  miles. 

4.  A  drover  bought  127  head  of  cattle  at  $34  a  head,  and  97 
head  at  $47  a  head,  and  sold  the  whole  lot  at  $40  a  head ;  what 
was  his  entire  profit  or  loss  ?  Ans.  $83  profit. 

5.  Multiply  675 —  (77  +  56)  by  (3  x  156)  — (214  — 28). 

Ans.  152844. 

6.  Multiply  98  ~\  6  x  (37  +  50)  by  (64  —  50;  x  5  — 10. 

Am.  37200. 

7.  What  is  the  product  of  (14  x  25)—  (9  x  36)  +  4324  x 
(280  —  112)  -f  (376~+  42)  x  4  ?  Ans    8,00-4,000. 

8.  In  1850  South  Carolina  cultivated  29967  farms  and  planta- 
tions, containing  an  average  of  541  acres  each,  at  an  average  value 
of  $2751  for  each  farm;  New  Jersey  cultivated  23905  farms,  con- 
taining an  average  of  115  acres  each,  at  an    average  value    of 
$5030  per  farm.     How  much  more  were  the  farming  lands  of  the 
latter  valued  at,  than  those  of  the  former  ? 

9.  There  are  in  the  United  States  1922890880  acres  of  land  ; 
of  this  there  were  reported  under  cultivation,  in  1850,  1449075 
farms,  each  embracing  an  average  of  203  acres.    How  many  acres 
were  still  uncultivated  ? 

10.  Each  of  the  above  farms  in  the  United  States  was  valued 
at  an  average  of  $2258,  and  upon  each  farm  there  was  an  average 


46  SIMPLE  NUMBERS. 

of  $105  in  implements  and  machinery.    What  was  the  aggregate 
value  of  the  farms  and  implements  ?          Ans.  $3,424,164,225. 
Find  the  values  of  the  following  expressions  : 

11.  24x  55  — 73?  Ans.  49,657. 

12.  153  — (32  x  25)  4-  2082  —  9  x  24?  Ans.  46,207. 

13.  22  +  33  4-  4*  4-  55  4  66  ? 

14.  In  1852  Great  Britain  consumed  1200000  bales  of  American 
cotton ;  allowing  each  bale  to  contain  400  pounds,   what   was    its 
total  weight  ? 

15.  If  a  house  is  worth  $2450,  and  the  farm  on  which  it  stands 
6  times  as  much,  lacking  $500,  and  the  stock  on  the  farm  twice 
as  much  as  the  house,  what  is  the  value  of  the  whole  ? 

Ans.  $21550. 

16.  A  flour  merchant  bought  1500  barrels  of  flour  at  7  dollars 
a  barrel ;  he  sold  800  barrels  at  10  dollars  a  barrel,  and  the  re- 
mainder at  6  dollars  a  barrel.     How  much  was  his  gain  ? 

17.  A  man  invests  in  trade  at  one  time  $450,  at  another  $780, 
at  another  $1250,  and  at  another  $2275 ;   how  much  must  he  add 
to  these  sums,  that  the  amount  invested  by  him  shall  be  increased 
fourfold?  Ans.  $14,265. 

18.  At  the  commencement  of  the  year  1858  there  were  in  ope- 
ration in  the  United  States  35000  miles  of  telegraph;   allowing 
the  average  cost  to  be  $115  per  mile,  what  was  the  total  cost? 

19.  The  cost  of  the  Atlantic  Telegraph  Cable,  as   originally 
made,  was  as  follows;  2500  miles  at  $485  per  mile,  10  miles  deep- 
sea  cable  at  $1450  per  mile,  and  25  miles  shore  ends  at  $1250  per 
mile.     What  was  its  total  cost  ?  Ans.  $1,258,250. 

20.  For  the  year  ending  June  30,  1859,  there  were  coined  in 
the  United  States  1401944  double  eagles  valued  at  twenty  dollars 
each,  62990  eagles,  154555  half  eagles,  and  22059  three  dollar 
pieces ;  what  was  the  total  value  of  this  gold  coin? 

Ans.  $29,507,732. 


DIVISION.  47 


DIVISION. 

1  OS.  Division  is  the  process  of  finding  how  many  times  one 
number  is  contained  in  another. 

1O  4.   The  Dividend  is  the  number  to  be  divided. 

105.    The  Divisor  is  the  number  to  divide  by. 

1OO.  The  Quotient  is  the  result  obtained  by  the  process  of 
division. 

1®  7.  The  Reciprocal  of  a  number  is  1  divided  by  the  number. 
Thus,  the  reciprocal  of  15  is  1  —•  15,  or  y1^. 

NOTES. — 1.  When  the  dividend  does  not  contain  the  divisor  an  exact  number 
of  times,  the  part  of  the  dividend  left  is  called  the  Remainder,  which  must  be 
less  than  the  divisor. 

2.  As  the  remainder  is  always  a  part  of  the  dividend,  it  is  always  of  the  same 
name  or  kind. 

3.  When  there  is  no  remainder  the  division  is  said  to  be  exact. 

1O8.  The  method  of  dividing  any  number  by  another  depends 
upon  the  following  principles  : 

I.  Division  is  the  reverse  of  multiplication,  the  dividend  cor- 
responding to  the  product,  and  the  divisor  and  quotient  to  the 
factors. 

II.  If  all  the  parts  of  a  number  be  divided,  the  entire  number 
will  be  divided. 

Since  the  remainder  in  dividing  any  part  of  the  dividend  must 
be  less  than  the  divisor,  it  can  be  divided  only  by  being  expressed 
in  units  of  a  lower  order.  Hence, 

III.  The  operation  must  commence  with  the  units  of  the  high- 
est order. 

1.  Divide  2742  by  6. 

ANALYSIS.     We  write  the  divisor  at  the  left 

of  the  dividend,  separated  from  it  by  a  line. 

As  6  is  not  contained  in  2  thousands,  we  take 

457  Ans.          the  2  thousands  and  7  hundreds  together,  and 

proceed  thus ;  6  is  contained  in  27  hundreds 

4  hundred  times,  and  the  remainder  is  3  hundreds ;  we  write  4  in 
hundreds'  place  in  the  quotient,  and  unite  the  remainder,  3  hundreds, 


48  SIMPLE  NUMBERS. 

to  the  next  figure  of  the  dividend,  making  34  tens ;  then,  6  is  con- 
tained in  34  tens  5  tens  times,  and  the  remainder  is  4  tens  ;  writing 
5  tens  in  its  place  in  the  quotient,  we  unite  the  remainder  to  the  next 
figure  in  the  dividend,  making  42 ;  6  is  contained  in  42  units  7  times, 
and  there  is  no  remainder ;  writing  7  in  its  place  in  the  quotient,  we 
have  the  entire  quotient,  457. 

NOTE  1. — The  different  numbers  which  we  divide  in  obtaining  the  successive 
figures  of  the  quotient,  are  called  partial  dividends. 

2.  Divide  18149  by  56. 

ANALYSIS.     As  neither  1  nor  18 
OPERATION.  ...         ,  .     ,,      ,.  . 

will  contain  the  divisor,  we  take 

56  )  18149  (  324/ff  Ans.         three  figures,  igl,  for  the  first  par- 

•^Q  tial  dividend.     56  is  contained  in 

134  181  3  times,  and  a  remainder  ;  we 

write  the  3  as  the  first  figure  in 

229  the  quotient,  and  then   multiply 

224  the  divisor  by  this  quotient  figure ; 

Z  3  times  56  is  168,  which  subtracted 

from  181,  leaves  13  ;  to  this  re- 
mainder we  annex  or  bring  down  4,  the  next  figure  of  the  dividend, 
and  thus  form  134,  the  next  partial  dividend;  56  is  contained  in 
134  2  times,  and  a  remainder;  2  times  56  is  112,  which  subtracted 
from  134,  leaves  22 ;  to  this  remainder  we  bring  down  9,  the  last 
figure  of  the  dividend,  and  we  have  229,  the  last  partial  dividend ;  56 
is  contained  in  229  4  times,  and  a  remainder ;  4  times  56  is  224, 
which  subtracted  from  229,  gives  5,  the  final  remainder,  which  we 
write  in  the  quotient  with  the  divisor  below  it,  thus  completing  the 
division,  (35). 

NOTE  2. — When  the  multiplication  and  subtraction  are  performed  mentally, 
ns  in  the  first  example,  the  operation  is  called  Short  Division ;  but  when  the 
work  is  written  out  in  full,  as  in  the  second  example,  the  operation  is  called 
Long  Division.  The  principles  governing  the  two  methods  are  the  same. 

1O9.  From  these  principles  and  illustrations  we  derive  the 
following  general 

RULE.  I.  Beginning  at  the  left  hand,  take  for  the  first  partial 
dividend  the  fewest  figures  of  the  given  dividend  that  will  contain 
the  divisor  one  or  more  times  ;  find  how  many  times  the  divisor  is 
contained  in  this  partial  dividend,  and  write  the  result  in  the 
quotient;  multiply  the  divisor  ly  this  quotient  figure,  and  subtract 
the  product  from  the  partial  dividend  used. 


DIVISION.  49 

II.  To  the  remainder  bring  down  the  next  figure  of  the  dividend, 
with  which  proceed  as  before ;  and  thus  continue  till  all  the  figures 
of  the  dividend  have  been  divided. 

III.  If  the  division  is  not  exact,  place  the  final  remainder  in 
the  quotient,  and  write  the  divisor  underneath. 

HO.  PROOF.  There  are  two  principal  methods  of  proving 
division. 

1st.  By  multiplication. 

Multiply  the  divisor  and  quotient  together,  and  to  the  product 
add  the  remainder,  if  any ;  if  the  result  be  equal  to  the  dividend, 
the  work  is  correct.  (1O8,  I.) 

NOTE. — In  multiplication,  the  two  factors  are  given  to  find  the  product;  in 
division,  the  product  and  one  of  the  factors  are  given  to  find  the  other  factor. 

2d.  By  excess  of  9's. 

111.  Subtract  the  remainder,  if  any,  from  the  dividend,  and 
find  the  excess  of  9's  in  the  result.  Multiply  the  excess  of  9's  in 
the  divisor  by  the  excess  of  9's  in  the  quotient,  and  find  the  excess 
of  9's  in  the  product;  if  the  latter  excess  is  the  same  as  the 
former,  the  work  is  supposed  to  be  correct.  (85.) 

EXAMPLES   FOR  PRACTICE. 

(1.)        (2.)        (3.)         (4.) 
6)473832   8)972496   9)1370961    12)73042164 

Quotients. 

5.  Divide  170352  by  36.  4732. 

6.  Divide  409887  by  47.  8721. 

7.  Divide  443520  by  84.  5280. 

8.  Divide  36380250  by  125.  291042. 

9.  Divide  1554768  by  216. 

10.  Divide  3931476  by  556. 

11.  Divide  48288058  by  3094.  Rem. 

12.  Divide  11214887  by  232.  7. 

13.  Divide  27085946  by  216.  194. 

14.  Divide  29137062  by  5317.  5219. 

15.  Divide  4917968967  by  2359.  1255- 
5                D 


50  SIMPLE  NUMBERS. 

16.  What  is  the  value  of  721198  -5-  291  ?  Rem.  100. 

17.  What  is  the  value  of  3844449-^-657?  342. 

18.  What  is  the  value  of  536819237  -h  907  ?  403. 

19.  What  is  the  value  of  571943007145  -~  37149  ?  12214. 

20.  What  is  the  value  of  48659910-^-54001  ?  5009. 

21.  The    annual    receipts   of   a    manufacturing    company  are 
$147675;  how  much  is  that  per  day,  there  being  365  days  in  the 
jear?  Arts.  $404|Jf. 

22.  The  New  York  Central  Railroad  Company,  in  1859,  owned 
556  miles  in  length  of  railroad,  which  cost,  for  construction  and 
equipment,  $30732518;  what  was  the  average  cost  per  mile? 

Ans.  $55,274Jf4. 

23.  The   Memphis  and  Charleston  Railroad  is  287  miles  in 
length,  and  cost  $5572470 ;  what  was  the  average  cost  per  mile  ? 

An*.  $19,416^V 

24.  The  whole  number  of  Post  offices  in  the  United  States,  in 
1858,  was  27977,  and  the  revenue  was  $8186793 ;  what  was  the 
average  income  to  an  office? 

ABBREVIATED    LONG   DIVISION. 

llSJo  We  may  avoid  writing  the  products  in  long  division, 
and  obtain  the  successive  remainders  by  the  method  of  subtraction 
employed  in  the  case  of  several  subtrahends.  (7G.) 

1.  Divide  261249  by  487. 

OPERATION.  ANALYSIS.     Dividing  the  first  partial 

487  )  261249  (  536          dividend,  2612,  we  obtain  5  for  the  first 

177  figure  of  the  quotient.     We  now  multi- 

313  ply  487  by  5  ;  but  instead  of  writing  the 

217  Kem.  product,  and   subtracting   it  from  the 

partial    dividend,  we    simply  observe 

what  figures  must  be  added  to  the  figures  of  the  product,  as  we  proceed, 
to  give  the  figures  of  the  partial  dividend,  and  write  them  for  the 
remainder  sought.  Thus,  5  times  7  are  35,  and  7  (written  in  the 
remainder,)  are  42,  a  number  whose  unit  figure  is  the  same  as  the 
right  hand  figure  of  the  partial  dividend ;  5  times  8  are  40,  and  4,  the 
tens  of  the  42,  are  44,  and  7  (written  in  the  remainder,)  are  51 ;  5 


DIVISION.  51 

times  4  are  20,  and  5,  the  tens  of  the  51,  are  25,  and  1  (written  in  the 
remainder,)  are  26.  We  next  consider  the  whole  rejnainder,  177,  as 
joined  with  4,  the  next  figure  of  the  dividend,  making  1774  for  the 
next  partial  dividend.  *  Proceeding  as  before,  we  obtain  313  for  the 
second  remainder,  217  for  the  final  remainder,  and  536  for  the  entire 
quotient.  Hence,  the  following 

RULE.     I.  Obtain  the  first  figure  in  the  quotient  in  the  usual 
manner. 

II.  Multiply  the  first  figure  of  the  divisor  by  this  quotient  figure, 
and  write  such  a  figure  in  the  remainder  as,  added  to  this  partial 
product,  will  give  an  amount  having  for  its  unit  figure  the  first  or 
right  hand  figure  of  the  partial  dividend  used. 

III.  Carry  the  tens'  figure  of  the  amount  to  the  product  of  the 
next  figure  of  the  divisor,  and  proceed  as  before,  till  the  entire 
remainder  is  obtained. 

IV.  Conceive  this  remainder  to  be  joined  to  the  next  figure  of 
the  dividend,  for  a  new  partial  dividend,  and  proceed  as  with  the 
former,  till  the  work  is  finished. 

EXAMPLES  FOR  PRACTICE. 

1.  Divide  77112  by  204.  Ans.  378. 

2.  Divide  65664  by  72.  Ans.  912. 

3.  Divide  7913576  by  209.  Ans.  37864. 

4.  Divide  6636584  by  698. 

5.  Divide  4024156  by  8903.  Ans.  452. 

6.  Divide  760592  by  6791. 

7.  Divide  101443929  by  25203.  Ans. 

8.  Divide  1246038849  by  269181.  Ans.  4629. 

9.  Divide  2318922  by  56240. 

10.  Divide  1454900  by  17300.          Ans. 


GENERAL  PRINCIPLES  OF  DIVISION. 

11*1.  The  general  principles  of  division  most  important  in  their 
application,  relate;  1st,  to  changing  the  terms  of  division  by 
addition  or  subtraction  ;  2d,  to  changing  the  terms  of  division 
by  multiplication  or  division  ;  3d,  to  successive  division. 


52 


SIMPLE  NUMBERS. 


11/4.  The  quotient  in  division  depends  upon  the  relative  values 
of  the  dividend  and  divisor  Hence,  any  change  in  the  value  of 
either  dividend  or  divisor  must  produce  a.change  in  the  value  of 
the  quotient;  though  certain  changes  may  be  made  in  both  divi- 
dend and  divisor,  at  the  same  time,  that  will  not  affect  the  quotient. 

CHANGING    THE    TERMS    BY   ADDITION    OR   SUBTRACTION. 

115.  Since  the  dividend  corresponds  to  a  product,  of  which 
the  divisor  and  quotient  are  factors,  we  observe, 

1st.  If  the  divisor  be  increased  by  1,  the  dividend  must  be 
increased  by  as  many  units  as  there  are  in  the  quotient,  in  order 
that  the  quotient  may  remain  the  same,  (03,  I) ;  and  if  the  divi- 
dend be  not  thus  increased,  the  quotient  will  be  diminished  by  as 
many  units  as  the  number  of  times  the  new  divisor  is  contained 
in  the  quotient.  Thus, 

84  -*-  6  =  14 

84  -*-  7  =  14  —  V  =  12 

2d.  If  the  divisor  be  diminished  by  1 ,  the  dividend  must  be 
diminished  by  as  many  units  as  there  are  in  the  quotient,  in  order 
that  the  quotient  may  remain  the  same,  (93,  II) ;  and  if  the 
dividend  be  not  thus  diminished,  the  quotient  will  be  increased 
by  as  many  units  as  the  number  of  times  the  new  divisor  is  con- 
tained in  the  quotient.  Thus, 

144  _=_  9  =  16 

144  _j_  8  =  16  +  V  =  18 

These  principles  may  be  stated  as  follows : 

I.  Adding  1  to  the  divisor  taJces  as  many  units  from  the  quotient 
as  the  new  divisor  is  contained  times  in  the  quotient. 

II.  Subtracting  1  from  the  divisor  adds  as  many  units  to  the 
quotient    as   the  new  divisor  is   contained    times   in   the   quotient. 
Hence, 

III.  ADDING  any  number  to  the  divisor  SUBTRACTS  as  many  units 
from  the  quotient  as  the  new  divisor  is  contained  times  in  the  pro- 
duct of  the  quotient  by  the  number  added;   and  SUBTRACTING 


DIVISION.  53 

any  number  from  the  divisor  ADDS  as  many  units  to  the  quotient 
as  the  new  divisor  is  contained  times  in  the  product  of  the  quo- 
tient by  the  number  subtracted. 

CHANGING   THE   TERMS   BY    MULTIPLICATION   OR   DIVISION. 

11O.    There  are  six  cases : 

1st.  If  any  divisor  is  contained  in  a  given  dividend  a  certain 
number  of  times,  the  same  divisor  will  be  contained  in  twice  the 
dividend  twice  as  many  times;  in  three  times  the  dividend,  three 
times  as  many  times  ;  and  so  on.  Hence, 

Multiplying  the  dividend  l>y  any  number,  multiplies  the  quotient, 
by  the  same  number. 

2d.  If  any  divisor  is  contained  in  a  given  dividend  a  certain 
number  of  times,  the  same  divisor  will  be  contained  in  one  half 
the  dividend  one  half  as  many  times ;  in  one  third  the  dividend, 
one  third  as  many  times ;  and  so  on.  Hence, 

Dividing  the  dividend  by  any  number,  divides  the  quotient  by  the 
same  number. 

3d.  If  a  given  divisor  is  contained  in  any  dividend  a  certain 
number  of  times,  twice  the  divisor  will  be  contained  in  the  same 
dividend  one  half  as  many  times  ]  three  times  the  divisor,  one 
third  as  many  times ;  and  so  on.  Hence, 

Multiplying  the  divisor  by  any  number,  divides  the  quotient  by 
the  same  number. 

4th.  If  a  given  divisor  is  contained  in  any  dividend  a  certain 
number  of  times,  one  half  the  divisor  will  be  contained  in  the 
same  dividend  twice  as  many  times ;  one  third  of  the  divisor,  three 
times  as  many  times ',  and  so  on.  Hence, 

Dividing  the  divisor  by  any  number,  multiplies  the  quotient  by 
the  same  number. 

5th.  It  a  given  divisor  is  contained  in  a  given  dividend  a  cer- 
tain number  of  times,  twice  the  divisor  will  be  contained  the  same 
number  of  times  in  twice  the  dividend ;  three  times  the  divisor 
will  be  contained  the  same  number  of  times  in  three  times  the 
dividend  ;  and  so  on.  Hence, 
5* 


54  SIMPLE  NUMBERS. 

Multiplying  both  dividend  and  divisor  by  the  same  number  does 
not  alter  the  quotient. 

6th.  If  a  given  divisor  is  contained  in  a  given  dividend  a  cer- 
tain number  of  times,  one  half  the  divisor  will  be  contained  the 
same  number  of  times  in  one  half  the  dividend  ;  one  third  of  the 
divisor  will  be  contained  the  same  number  of  times  in  one  third 
of  the  dividend  ;  and  so  on.  Hence, 

Dividing  both  dividend  and  divisor  by  the  same  number  does  not 
alter  the  quotient. 

NOTE.  —  If  a  number  be  multiplied  and  the  product  divided  by  the  same  num- 
ber, the  quotient  will  be  equal  to  the  number  multiplied;  hence  the  5th  case 
may  be  regarded  as  a  direct  consequence  of  the  1st  and  3d;  and  the  6th,  as  the 
direct  consequence  of  the  2d  and  4th. 

To  illustrate  these  cases,  take  24  for  a  dividend  and  6  for  a 
divisor  •  then  the  quotient  will  be  4,  and  the  several  changes  may 
be  represented  in  theii  order  as  follows  : 

Dividend.      Divisor.     Quotient. 

24    -*-    6    =    4 


i      4%  n  o  |  Multiplying  the  dividend  by  2  multi- 

8  {      plies  the  quotient  by  2. 

n     in  a  of  Dividing   the   dividend  by   2  divides 

1  \      the  quotient  by  2. 

o     9  •          -19  of  Multiplying  the  divisor  by  2  divides 

-  {      the  quotient  by  2. 

Q  f  Dividing   the  divisor  by  2   multiplies 
8  {      the  quotient  by  2. 

,    -19  .  (  Multiplying  both  dividend  and  divisor 

{      by  2  does  not  alter  the  quotient. 

fi      1  9  o  A  f  Dividing  both  dividend  and  divisor  by 

{      2  does  not  alter  the  quotient. 

117.  These  six  cases  constitute  three  general  principles, 
which  may  now  be  stated  as  follows  : 

PRIN.  I.  Multiplying  the  dividend  multiplies  the  quotient;  and 
dividing  the  dividend  divides  the  quotient. 

PRIN.  II.  Multiplying  the  divisor  divides  the  quotient;  and 
dividing  the  divisor  multiplies  the  quotient. 


DIVISION.  55 

PRIN.  III.  Multiplying  or  dividing  both  dividend  and  divisor 
by  the  same  number,  does  not  alter  the  quotient. 

118.   These  three  principles  may  be  embraced  in  one 

GENERAL    LAW. 

A  change  in  the  dividend  produce*  a  LIKE  change  in  the  quo- 
tient; but  a  change  in  the  divisor  produces  an  OPPOSITE  change 
in  the  quotient. 

SUCCESSIVE  DIVISION. 

11O.  Successive  Division  is  the  process  of  dividing  one 
number  by  another,  and  the  resulting  quotient  by  a  second  divisor, 
and  so  on. 

Successive  division  is  the  reverse  of  continued  multiplication. 
Hence, 

I.  If  a  given  number  be  divided  by  several  numbers  in  succes- 
sive division,  the  result  will  be  the  same  as  if  the  given  number 
were  divided  by  the  product  of  the  several  divisors,  (O«5,  I). 

II.  The  result  of  successive  division  is  the  same,  in  whatever 
order  the  divisors  are  taken,  (95,  II). 

CONTRACTIONS  IN  DIVISION. 
CASE   I. 

12O.   When  the  divisor  is  a  composite  number. 
1.  Divide  1242  by  54. 

OPERATION  ANALYSIS.     The  component  factors  of  54  are 

6")  1?49  6  and  9.     We  divide  1242  by  6,  and  the  re- 

sulting quotient  by  9,  and  obtain  for  the  final 
result,  23,  which  must  be  the  same  as  the 
23  Ans.         quotient  of  1242  divided  by  6  times  9,  or  54, 
(119,  I).     We  might  have  obtained  the  same 

result  by  dividing  first  by  9,  and  then  by  6,  (119,  II).      Hence  the 
following 

RULE.     Divide  the  dividend  by  one  of  the  factors,  and  the  quo* 


56  SIMPLE  NUMBERS. 

tient  thus  obtained  by  another,  and  so  on  if  there  be  more  than  two 
factors,  until  every  factor  has  been  made  a  divisor.  The  last  quo- 
tient will  be  the  quotient  required. 


TO    FIND    THE    TRUE    REMAINDER. 

If  remainders  occur  in  successive  division,  it  is  evident 
that  the  true  remainder  must  be  the  least  number,  which,  sub- 
tracted from  the  given  dividend,  will  render  all  the  divisions 
exact 

1.  Divide  5855  by  168,  using  the  factors  3,  7,  and  8,  and  find 
the  true  remainder. 

OPERATION.  ANALYSIS.      Dividing    the 

3)  5855  given  dividend  by  3,  we  have 


7)1951  2         "          or  a 

-  -  remainder   of  2.     Hence,   2 

8)2'8  .........  5x3=    15          subtracted  from  5855  would 

34  ...  6  X  7  X  3  =126          render  the  first  division  exact, 
True  remainder  ...........................  143          and  we  therefore  write  2  for 

a  part  of  the  true  remainder. 

Dividing  1951  by  7,  we  have  278  for  a  quotient,  and  a  remainder  of  5. 
Hence,  5  subtracted  from  1951  would  render  the  second  division  exact. 
But  to  dimmish  1951  by  5  would  require  us  to  diminish  1951  X  3,  the 
dividend  of  the  first  exact  division,  by  5  X  3  —  15,  (93,  III)  ;  and 
we  therefore  write  15  for  the  second  part  of  the  true  remainder. 
Dividing  278  by  8,  we  have  34  for  a  quotient,  and  a  remainder  of  6. 
Hence,  6  subtracted  from  278  would  render  the  third  division  exact. 
But  to  diminish  278  by  6  would  require  us  to  diminish  278  X  7,  the 
dividend  of  the  second  exact  division,  by  6  X  7  ;  or  278  X  7  X  3,  the 
dividend  of  the  first  exact  division,  by  6  X  7  X  3  =  126  ;  and  we 
therefore  write  126  for  the  third  part  of  the  true  remainder.  Adding 
the  three  parts,  we  have  143  for  the  entire  remainder. 

Hence  the  following 

RULE.  I.  Multiply  each  partial  remainder  by  all  the  preceding 
divisors. 

II.  Add  the  several  products;  the  sum  will  be  the  true  re- 
mainder. 


DIVISION.  57 

EXAMPLES  FOR  PRACTICE. 

1.  Divide  435  by  15  =  3  x  5.  Ans.  29. 

2.  Divide  4256  by  56  =  7  X  8. 

3.  Divide  17856  by  72  =  9  x  8. 

4.  Divide  15288  by  42  =  2  x  3  x  7.  Am.  364. 

5.  Divide  972552  by  168  =  8  x  7  x  3.  Ans.  5789. 

6.  Divide  526050  by  126  =  9  x  7  X  2. 

7.  Divide  612360  by  105  =  7  x  5  x  3.  Ans.  5832. 

8.  Divide  553  by  15  =  3  x  5.  Rem.  13. 

9.  Divide  10183  by  105  =  3  x  5  x  7.          103. 

10.  Divide  10197  by  120  =  2  x  3  x  4  x  5.  117. 

11.  Divide  29792  by  144  =  3  x  8  x  6.  128. 

12.  Divide  73522  by  168  =  4  x  6  X  7.  106. 

13.  Divide  63844  by  135  =  3  x  5  x  9,  124. 

14.  Divide  386639  by  720  =  2  x  3  x  4  x  5  x  6.  719. 

15.  Divide  734514  by  168  =  4  x  6  X  7.  18. 

16.  Divide  636388  by  729  =  9s.  700. 

17.  Divide  4619  by  125  =  53.  119. 

18.  Divide  116423  by  10584  =  3  x  72  x  8  x  9.  10583. 

19.  Divide  79500  by  6125  =  53  x  72.  6000. 

CASE  II. 

122.   When  the  divisor  is  a  unit  of  any  order. 

If  we  cut  off  or  remove  the  right  hand  figure  of  a  number,  each 
of  the  other  figures  is  removed  one  place  toward  the  right,  and, 
consequently,  the  value  of  each  is  diminished  tenfold,  or  divided 
by  10,  (oT,  III).  For  a  similar  reason,  by  cutting  off  two  figures 
we  divide  by  100 ;  by  cutting  off  three,  we  divide  by  1000,  and 
so  on;  and  the  figures  cut  off  will  constitute  the  remainder. 
Hence  the 

RULE,  from  the  right  hand  of  the  dividend  cut  off  as  many 
figures  as  there  are  ciphers  in  the  divisor.  Under  the  figures  so 
cut  off,  place  the  divisor,  and  the  whole  will  form  the  quotient. 


58  SIMPLE  NUMBERS. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  79  by  10.  Ans.  7T% 

2.  Divide  7982  by  100. 

3.  Divide  4003  by  1000.  Ans. 

4.  Divide  2301050  by  10000. 

5.  Divide  3600036  by  1000.  Ans. 

CASE    III. 

123.   When  there  are  ciphers  on  the  right  hand  of 
the  divisor. 

I.  Divide  25548  by  700. 

OPERATION.  ANALYSIS.     We  resolve  700 

into  the   factors   10°   and  7- 
Dividing  first  by  100,  the  quo- 

36  Quotient.    3  2<i  rem  tient  is  255,  and  the  remainder 

3  X  100  +  48  =  348  true  rem.       48.     Dividing  255  by  7,  the 

final  quotient  is  36,  and  the 

second  remainder  3.  Multiplying  the  last  remainder,  3,  by  the 
preceding  divisor,  100,  and  adding  the  preceding  remainder,  we  have 
300+48  =  348,  the  true  remainder,  (121).  In  practice,  the  true 
remainder  may  be  obtained  by  prefixing  the  second  remainder  to 
the  first.  Hence  the 

HULE.     I.    Cut  off  the  ciphers  from  the  right  of  the  divisor,  and 
as  many  figures  from  the  right  of  the  dividend. 

II.  Divide  the  remaining  figures  of  the  dividend  by  the  remain- 
ing figures  of  the  divisor,  for  the  final  quotient. 

III.  Prefix  the  remainder  to  the  figures  cut  off,  and  the  result 
toitt  be  the  true  remainder. 

EXAMPLES  FOR  PRACTICE. 

1.  Divide  7856  by  900.  Ans.  8«|>g. 

2.  Divide  13872  by%  500. 

3.  Divide  83248  by  2600.  Ans. 

4.  Divide  1548036  by  4300.  Ans. 

5.  Divide  436000  by  300.  Ans. 

6.  Divide  66472000  by  8100. 
1.  Divide  10818000  by  3600. 


DIVISION.  59 


EXAMPLES    COMBINING   THE    PRECEDING    RULES. 

1.  How  many  barrels  of  flour  at  $8  a  barrel,  will  pay  for  25  tons 
of  coal  at  $4  a  ton,  and  36  cords  of  wood  at  $3  a  cord  ? 

Ans.  26. 

2.  A  grocer  bought  12  barrels  of  sugar  at  $16  per  barrel,  and 
17  barrels  at  $13  per  barrel ;  how  much  would  he  gain  by  selling 
the  whole  at  $18  per  barrel  ? 

3.  A  farmer  sold  300  bushels  of  wheat  at  $2  a  bushel,  corn  and 
oats  to  the  amount  of  $750 ;  with  the  proceeds  he  bought  120 
head  of  sheep  at  $3  a  head,  one  pair  of  oxen  for  $90,  and  25  acres 
of  land  for  the  remainder      How  much  did  the  land  cost  him  per 
acre  ? Ans.  $36. 

4.  Divide  450 -f  (24  — 12)  x  5  by  (90  -i-  6)  -f  (3  x  11)  — 18. 

Ans.  17. 

5.  Divide  648  x  (32  x  23)  -*-  9  —  (2910  -«-  15)  by  2863  ~ 
(4375  -TT75)  X  42  +  32 .  Ans.  712f . 

6.  The   product  of  three   numbers   is  107100;    one  of    the 
numbers  is  42,  and  another  34.     What  is  the  third  number  ? 

Ans.  75. 

7.  What  number  is  that  which  being  divided  by  45,  the  quo- 
tient increased  by  72  -f  1>  the  sum  diminished  by  the  difference 
between  28  and  16,  the  remainder  multiplied  by  6,  and  the  pro- 
duct divided  by  24,  the  quotient  will  be  12  ?  Ans.  450. 

8.  A  mechanic  earns  $60  a  month,  but  his  necessary  expenses 
are  $42  a  month.     How  long  will  it  take  him  to  pay  for  a  farm 
of  50  acres  worth  $36  an  acre  ? 

9.  What  number  besides  472  will  divide  251104  without  a  re- 
mainder? Ans.  532. 

10.  Of  what  number  is  3042  both  divisor  and  quotient  ? 

Ans.  9253764. 

11.  What  must  the  number  be  which,  divided  by  453,  will  give 
the  quotient  307,  and  the  remainder  109  ?  Ans.  139180. 

12.  A  farmer  bought  a  lot  of  sheep  and  hogs,  of  each  an  equal 
number,  for  $1276.     He  gave  $4  a  head  for  the  sheep,  and  $7  a 


50  SIMPLE  NUMBERS 

head  for  the  hogs ;  what  was  the  whole  number  purchased,  and 
how  much  was  the  difference  in  the  total  cost  of  each  ? 

Ans.  232  purchased ;  $348  difference  in  cost. 

13.  According  to  the  census  of  1850  the  total  value  of  the 
tobacco  raised  in  the  United  States  was  $13,982,686.     How  many 
school-houses  at  a  cost  of  $950,  and  churches  at  a  cost  of  $7500, 
of  each  an  equal  number,  could  be  built  with  the  proceeds  of  the 
tobacco  crop  of  1850  ?       Ans.  1654,  and  a  remainder  of  $6386. 

14.  The  entire  cotton  crop  in  the  United  States  in  1859  was 
4,300,000  bales,  valued  at  $54  per  bale.     If  the  entire  proceeds 
were  exchanged  for  English  iron,  at  $60  per  ton,  how  many  tons 
would  be  received  ? 

15.  The  population  of  the  United  States  in  1850  was  23,191,876. 
It  was  estimated  that  1  person  in  every  400  died  of  intemperance. 
How  many  deaths  may  be  attributed  to  this  cause  in  the  United 
States,  during  that  year  ? 

16.  In  1850,  there  were  in  the  State  of  New  York,  10,593 
public  schools,  which  were  attended  during  the  winter  by  508464 
pupils ;  what  was  the  average  number  to  each  school  ? 

Ans.  48. 

17.  A  drover  bought  a  certain  number  of  cattle  for  $9800,  and 
sold  a  certain  number  of  them  for  $7680,  at  $64  a  head,  and 
gained  on  those  he  sold  $960.     How  much  did  he  gain  a  head, 
and  how  many  did  he  buy  at  first  ? 

Ans.  Gained  $8  per  head;  bought  175. 

18.  A  house  and  lot  valued  at  $1200,  and  6  horses  at  $95  each, 
were  exchanged  for  30  acres  of  land.     At  how  much  was  the  land 
valued  per  acre  ? 

19.  If  16  men  can  perform  a  job  of  work  in  36  days,  in  how 
many  days  can  they  perform  the  same  job  with  the  assistance  of 
8  more  men  ?  Ans.  24. 

20.  Bought  275  barrels  of  flour  for  $1650,  and  sold  186  bar- 
rels of  it  at  $9  a  barrel,  and  the  remainder  for  what  it  cost.    How 
much  was  gained  by  the  bargain  ?  Ans.  $558. 

21    A  grocer  wishes  to  put  840  pounds  of  tea  into  three  kinds 
of  boxes,  containing  respectively  5,  10,  and  15  pounds,  using  the 


PROBLEMS.  61 

same  number  of  boxes  of  each  kind.     How  many  boxes  can  he 
fill?  Ans.  84. 

22.  A  coal  dealer  paid  $965  for  some  coal.     He  sold  160  tons 
for  $5  a  ton,  when  the  remainder  stood  him  in  but  $3  a  ton.     How 
many  tons  did  he  buy  ?  Ans.  215. 

23.  A  dealer  in  horses  gave  $7560  for  a  certain  number,  and 
sold  a  part  of  them  for  $3825,  at  $85  each,  and  by  so  doing,  lost 
$5  a  head ;  for  how  much  a  head  must  he  sell  the  remainder,  to 
gain  $945  on  the  whole  ?  Ans.  $120. 

24.  Bought  a  Western  farm  for  $22,360,  and  after  expending 
$1742  in  improvements  upon  it,  I  sold  one  half  of  it  for  $15480, 
at  $18  per  acre.     How  many  acres  of  land  did  I  purchase,  and  at 
what  price  per  acre  ? 

PKOBLEMS  IN  SIMPLE  INTEGRAL  NUMBERS. 

124.  The  four  operations  that  have  now  been  considered,  viz., 
Addition,  Subtraction,  Multiplication,  and  Division,  are  all  the 
operations  that  can  be  performed  upon  numbers,  and  hence  they 
are  called  the  Fundamental  Rules. 

12o.  In  all  cases,  the  numbers  operated  upon  and  the  results 
obtained,  sustain  to  each  other  the  relation  of  a  whole  to  its  parts. 
Thus, 

I.  In  Addition,  the  numbers  added  are  the  parts,  and  the  sum 

or  amount  is  the  whole. 

II.  In   Subtraction,  the    subtrahend    and    remainder  are  the 
parts,  and  the  minuend  is  the  whole. 

III.  In  Multiplication ,  the  multiplicand  denotes  the  value  of  one 

part,  the  multiplier  the  number  of  parts,  and  the  pro- 
duct the  total  value  of  the  whole  number  of  parts. 

IV.  In  Division,  the  dividend  denotes  the  total  value  of  the 

whole  number  of  parts,  the  divisor  the  value  of  one 
part,  and  the  quotient  the  number  of  parts ;  or  the 
divisor  the  number  of  parts,  and  the  quotient  the 
value  of  one  part. 

126.  Every  example  that  can  possibly  occur  in  Arithmetic, 
and  every  business  computation  requiring  an  arithmetical  opera- 


^        °r 


62  SIMPLE  NUMBERS. 

tion,  can  be  classed  under  one  or  more  of  the  four  Fundamental 
Rules,  as  follows  : 

I.  Cases  requiring  Addition. 
There  may  Le  given  To  find 

1.  The  parts,  the  whole,  or  the  sum  total. 

2  The  less  of  two  numbers  and 
their  difference,  or  the  sub- 
trahend  and  remainder, 

II.  Cases  requiring  Subtraction. 
There  may  be  <jiven  To  find 

1.  The  sum  of  two  numbers  and  ") 

one  of  them,  j  the  other- 

2.  The  greater  and  the  less  of  ^ 

two  numbers,  or  the  minuend  I  the  difference  or  remainder 
and  subtrahend,  J 

III.  Cases  requiring  Multiplication. 
There  may  le  given  To  find 

1.  Two  numbers,  their  product. 

2.  Any  number  of  factors,  their  continued  product. 

3.  The  divisor  and  quotient,  the  dividend. 

IV.  Cases  requiring  Division. 
There  may  le  given  To  find 

1.  The  dividend  and  divisor,          the  quotient. 

2.  The  dividend  and  quotient,       the  divisor. 

8.  The  product  and  one  of  two  ) 

„    .  }•  the  other  factor. 

factors,  ) 

4.  The  continued    product   of  ^ 

several  factors,  and  the  pro-  I  that  one  factor. 
duct  of  all  but  one  factor,     J 

ISBT.   Let  the  pupil  be  required  to  illustrate  the  following  pro- 
blems by  original  examples. 

Problem  1.  Given,  several  numbers,  to  find  their  sum. 
Prob.  2.  Given,  the  sum  of  several  numbers  and  all  of  them 
but  one,  to  find  that  one. 


PROBLEMS.  63 

Prob.  3.  Given,  the  parts,  to  find  the  whole. 

Prob.  4.  Given,  the  whole  and  all  the  parts  but  one,  to  find 
that  one. 

Prob.  5.  Given,  two  numbers,  to  find  their  difference. 

Prob.  6.  Given,  the  greater  of  two  numbers  and  their  difference, 
to  find  the  less  number. 

Prob.  7.  Given,  the  less  of  two  numbers  and  their  difference,  to 
find  the  greater  number. 

Prob.  8.  Given,  the  minuend  and  subtrahend,  to  find  the 
remainder. 

Prob.  9.  Given,  the  minuend  and  remainder,  to  find  the  sub- 
trahend. 

Prob.  10.  Given,  the  subtrahend  and  remainder,  to  find  the 
minuend. 

Prob.  11.  Given,  two  or  more  numbers,  to  find  their  product. 

Prob.  12.  Given,  the  product  and  one  of  two  factors,  to  find  the 
other  factor. 

Prob.  13.  Given,  the  continued  product  of  several  factors  and 
all  the  factors  but  one,  to  find  that  factor. 

Prob.  14.  Given,  the  factors,  to  find  their  product. 

Prob.  15  Given,  the  multiplicand  and  multiplier,  to  find  the 
product. 

Prob.  16.  Given,  the  product  and  multiplicand,  to  find  the 
multiplier. 

Prob.  17.  Given,  the  product  and  multiplier,  to  find  the  mul- 
tiplicand. 

Prob.  18    Given,  two  numbers,  to  find  their  quotients. 

Prob.  19.  Given,  the  divisor  and  dividend,  to  find  the  quotient. 

Prob.  20.  Given,  the  divisor  and  quotient,  to  find  the  dividend. 

Prob.  21.  Given,  the  dividend  and  quotient,  to  find  the  divisor. 

Prob.  22.  Given,  the  divisor,  quotient,  and  remainder,  to  find 
the  dividend. 

Prob.  23.  Given,  the  dividend,  quotient,  and  remainder,  to  find 
the  divisor. 

Prob  24.  Given,  the  final  quotient  of  a  continued  division  and 
the  several  divisors,  to  find  the  dividend. 


64  SIMPLE  NUMBERS, 

« 

Prob  25.  Given,  the  final  quotient  of  a  continued  division,  the 
first  dividend,  and  all  the  divisors  but  one,  to  find  that  divisor. 

Prob.  26.  Given,  the  dividend  and  several  divisors  of  a  con- 
tinued division,  to  find  the  quotient. 

Prob.  27.  Given,  two  or  more  sets  of  numbers,  to  find  the 
difference  of  their  sums. 

Prob.  28.  Given,  two  or  more  sets  of  factors,  to  find  the  sum  of 
their  products. 

Prob.  29.  Given,  one  or  more  sets  of  factors  and  one  or  more 
numbers,  to  find  the  sum  of  the  products  and  the  given  numbers. 

Prob.  30.  Given,  two  or  more  sets  of  factors,  to  find  the  differ- 
ence of  thoir  products. 

Prob.  31.  Given,  one  or  more  sets  of  factors  and  one  or  more 
numbers,  to  find  the  sum  of  the  products  and  the  given  number 
or  numbers. 

Prob  32.  Given,  two  or  more  sets  of  factors  and  two  or  more 
other  sets  of  factors,  to  find  the  difference  of  the  sums  of  the 
products  of  the  former  and  latter. 

Prob  33.  Given,  the  sum  and  the  difference  of  two  numbers,  to 
find  the  numbers. 

ANALYSIS.  If  the  difference  of  two  unequal  numbers  be  added  to 
the  less  number,  the  sum  will  be  equal  to  the  greater ;  and  if  this 
sum  be  added  to  the  greater  number,  the  result  will  be  twice  the 
greater  number  But  this  result  is  the  sum  of  the  two  numbers  plus 
their  difference 

Again,  if  the  difference  of  two  numbers  be  subtracted  from  the 
greater  number,  the  remainder  will  be  equal  to  the  less  number  ;  and 
if  this  remainder  be  added  to  the  less  number,  the  result  will  be  twice 
the  less  number.  But  this  result  is  the  sum  of  the  two  numbers 
minus  their  difference.  Hence, 

I.  The  sum  of  two  numbers  plus  their  difference  is  equal  to 
twice  the  greater  number 

II.  The  sum  of  two  numbers  minus  their  difference  is  equal  to 
twice  the  less  number. 


EXACT  DIVISORS.  65 


PROPERTIES  OF  NUMBERS. 

EXACT  DIVISORS. 

An  Exact  Divisor  of  a  number  is  one  that  gives  an 
integral  number  for  a  quotient.  And  since  division  is  the  reverse 
of  multiplication,  it  follows  that  all  the  exact  divisors  of  a  number 
are  factors  of  that  number,  and  that  all  its  factors  are  exact 
divisors. 

NOTES.  —  1.  Every  number  is  divisible  by  itself  and  unity;  but  the  number 
itself  and  unity  are  not  generally  considered  as  factors,  or  exact  divisors  of  the 
number. 

2.  An  exact  divisor  of  a  number  is  sometimes  called  the  measure  of  the 
number. 


An  Even  Number  is  a  number  of  which  2  is  an  exact 
divisor;  as  2,  4,  6,  or  8. 

130.  An  Odd  Number  is  a  number  of  which  2  is  not  an  exact 
divisor;  as  1,  3,  5,  7,  or  9. 

131.  A  Perfect  Number  is  one  that  is  equal  to  the  sum  of 
all  its  factors  plus  1;  as  6  =  3  +  2  +  1,  or  28  =  14  +  7  +  4  + 
2  +  1. 

NOTE  —The  only  perfect  numbers  known  are  6,  28,  496,  8128,  33550336, 
8589869056,  137438691328,  2305843008139952128,  2417851639228158837784576, 
990352031428297183044881  6128. 


An  Imperfect  Number  is  one  that  is  not  equal  to  the 
sum  of  all  its  factors  plus  1  ,  as  12,  which  is  not  equal  to  6  -f  4 
+  3  +  2  +  1. 

133.  An  Abundant  Number  is  one  which  is  less  than  the 
sum  of  all  its  factors  plus  1  ;  as  18,  which  is  less  than  9  +  6  + 
3  +  2+1. 

134.  A  Defective  Number  is  one  which  is  greater  than  the 
sum  of  all  its  factors  plus  1  ;  as  27,  which  is  greater  than  9  +  3  +  1. 

135.  To  show  the  nature  of  exact  division,  and  furnish  tests 
of  divisibility,  observe  that  if  we  begin  with  any  number,  as  4, 
and  take  once  4,  two  times  4,  three  times'  4,  four  times  4,  and  so 
on  indefinitely,  forming  the  series  4,  8,  12,  16,  etc.,  we  shall  have 


66  PROPERTIES  OF  NUMBERS. 

all  the  numbers  that  are  divisible  by  4 ;  and  from  the  manner  of 
forming  this  series,  it  is  evident, 

1st.  That  the  product  of  any  one  number  of  the  series  by  any 
integral  number  whatever,  will  contain  4  an  exact  number  of 
times  'j 

2d.  The  sum  of  any  two  numbers  of  the  series  will  contain  4 
an  exact  number  of  times ;  and 

3d.  The  difference  of  any  two  will  contain  4  an  exact  number 
of  times.  Hence, 

I  Any  number  which  will  exactly  divide  one  of  two  numbers 
will  divide  their  product. 

II.  Any  number  which  will  exactly  divide  each  of  two  numbers 
will  divide  their  sum. 

III.  Any  number  which  will  exactly  divide  each  of  two  num- 
bers will  divide  their  difference. 

13O.    From  these  principles  we  derive  the  following  properties : 

I.  Any  number  terminating  with  0,  00,  000,  etc.,  is  divisible 
by  10,  100,  1000,  etc.,  or  by  any  factor  of  10,  100,  or  1000. 

For  by  cutting  off  the  cipher  or  ciphers,  the  number  will  be  divided 
by  10.  100.  or  1000,  etc.,  without  a  remainder,  (122) ;  and  a  number 
of  which  10,  100,  or  1000,  etc.,  is  a  factor,  will  contain  any  factor  of 
10,  100,  or  1000,  etc.,  (I). 

II.  A  number  is  divisible  by  2  if  its  right  hand  figure  is  even 
or  divisible  by  2. 

For,  the  part  at  the  left  of  the  units'  place,  taken  alone,  with  its 
local  value,  is  a  number  which  terminates  with  a  cipher,  and  is  divi- 
sible by  2,  because  2  is  a  factor  of  10,  (I) ;  and  if  both  parts,  taken 
separately,  with  their  local  values,  are  divisible  by  2,  their  sum,  which 
is  the  entire  number,  is  divisible  by  2,  (135,  II). 

NOTE. — Hence,  nil  numbers  terminating  with  0,  2,  4,  6,  or  8,  are  even,  and  all 
numbers  terminating  with  1,  3,  5,  7,  or  9,  are  odd. 

III.  A  number  is  divisible  by  4  if  the  number  expressed  by  its 
two  right  hand. figures  is  divisible  by  4. 

For,  the  part  at  the  left  of  the  tens'  place,  taken  alone,  with  its 
local  value,  is  a  number  which  terminates  with  two  ciphers,  and  is 
divisible  by  4,  because  4  is  a  factor  of  100,  (I)  ;  and  if  both  parts, 


EXACT  DIVISORS.  67 

taken  separately,  with  their  local  values,  are  divisible  by  4,  their  sum 
which  is  the  entire  number,  is  divisible  by  4,  (135,  II) 

IV.  A  number  is  divisible  by  8  if  the  number  expressed  by  its 
three  right  hand  figures  is  divisible  by  8. 

For,  the  part  at  the  left  of  the  hundreds'  place,  taken  alone,  with 
its  local  value,  is  a  number  which  terminates  with  three  ciphers,  and 
is  divisible  by  8,  because  8  is  a  factor  of  1000,  (I) ;  and  if  both 
parts,  taken  separately,  with  their  local  values,  are  divisible  by  8, 
their  sum,  or  the  entire  number,  ~s  divisible  by  8,  (135,  II). 

V.  A  number  is  divisible  by  any  power  of  2,  if  as  many  right 
hand  figures  of  the  number  as  are  equal  to  the  index  of  the  given 
power,  are  divisible  by  the  given  power. 

For,  as  2  is  a  factor  of  10,  any-  power  of  2  is  a  factor  of  the  corres- 
ponding power  of  10,  or  of  a  unit  of  an  order  one  higher  than  is 
indicated  by  the  index  of  the  given  power  of  2 ;  and  if  both  parts 
of  a  number,  taken  separately,  with  their  local  values,  are  divisible 
by  a  power  of  2,  their  sum,  or  the  entire  number,  is  divisible  by  the 
same  power  of  2,  (135,  II). 

VI.  A  number  is  divisive  by  5  if  its  right  hand  figure  is  0, 
or  5. 

For,  if  a  number  terminates  with  a  cipher,  it  is  divisible  by  5, 
because  5  is  a  factor  of  10,  (I)  ;  and  if  it  terminates  with  5,  both 
parts,  the  units  and  the  figures  at  the  left  of  units,  taken  separately, 
with  their  local  values,  are  divisible  by  5,  and  consequently  their 
sum,  or  the  entire  number,  is  divisible  by  5,  (135,  II). 

VII.  A  number  is  divisible  by  25  if  the  number  expressed  by 
its  two  right  hand  figures  is  divisible  by  25. 

For,  the  part  at  the  left  of  the  tens'  figure,  taken  with  its  local 
value,  is  a  number  terminating  with  two  ciphers,  and  is  divisible  by 
25,  because  25  Is  a  factor  of  100,  (I)  ;  and  if  both  parts,  taken 
separately,  with  their  local  values,  are  divisible  by  25,  their  sum,  or 
the  entire  number,  is  divisible  by  25,  (135,  II). 

VIII.  A  number  is  divisible  by  any  power  of  5,  if  as  many 
right  hand  figures  of  the  number  as  are  equal  to  the  index  of  the 
given  power  are  divisible  by  the  given  power. 

For,  as  5  is  a  factor  of  10,  any  power  of  5  is  a  factor  of  the  corres- 
ponding power  of  10,  or  of  a  unit  of  an  order  one  higher  than  is  indi- 


68  PROPERTIES  OF  NUMBERS. 

eated  by  the  index  of  the  given  power  of  5  ;  and  if  both  parts  of  a 
number,  taken  separately,  with  their  local  values,  are  divisible  by  a 
power  of  5,  their  sum,  or  the  entire  number,  is  divisible^by  the  same 
power  of  5,  (135,  II). 

IX.  A  number  is  divisible  by  9  if  the  sum  of  its  digits  is  divis- 
ible by  9. 

For,  if  any  number,  as  7245,  be  separated  into  its  parts,  7000  + 
200  +40+5,  and  each  part  be  divided  by  9,  the  several  remainders 
will  be  the  digits  7,  2,  4,  and  5,  respectively ;  hence,  if  the  sum  of 
these  digits,  or  remainders,  be  9  or  an  exact  number  of  9's,  the  entire 
number  must  contain  an  exact  number  of  9's,  and  will  therefore  be 
divisible  by  9. 

NOTE.  —  Whence  it  follows  that  if  a  number  be  divided  by  9,  the  remainder 
will  be  the  same  as  the  excess  of  9's  in  the  sum  of  the  digits  of  the  number. 
Upon  this  property  depends  one  of  the  methods  of  proving  the  operations  in 
the  four  Fundamental  Rules. 

X.  A  number  is  divisible  by  a  composite  number,  when  it  is 
divisible,  successively,  by  all  the  component  factors  of  the  com- 
posite number. 

For,  dividing  any  number  successively  by  several  factors,  is  the 
same  as  dividing  by  the  product  of  these  factors,  (119,  I). 

XI.  An  odd  number  is  not  divisible  by  an  even  number. 

For,  the  product  of  any  even  number  by  any  odd  number  is  even ; 
and,  consequently,  any  composite  odd  number  can  contain  only  odd 
factors. 

XII.  An  even  number  that  is  divisible  by  an  odd  number,  is 
also  divisible  by  twice  that  odd  number. 

For,  if  any  even  number  be  divided  by  an  odd  number,  the  quo- 
tient must  be  even,  and  divisible  by  2 ;  hence,  the  given  even  num- 
ber, being  divisible  successively  by  the  odd  number  and  2,  will  be 
divisible  by  their  product,  or  twice  the  odd  number,  (119,  I). 

PRIME  NUMBERS. 

137.  A  Prime  Number  is  one  that  can  not  be  resolved  or 
eeparated  into  two  or  more  integral  factors. 

NOTE.  —  Every  number  must  be  either  prime  or  composite. 


PRIME  NUMBERS.  69 

138.  To  find  all  the  prime  numbers  within  any  given  limit, 
we  observe  that  all  even  numbers  except  2  are  composite ;  hence, 
the  prime  numbers  must  be  sought  among  the  odd  numbers 

130,  If  the  odd  numbers  be  written  in  their  order,  thus ;  1, 
3,  5,  7,  9,  11,  13,  15  17,  etc.,  we  observe, 

1st.  Taking  every  third  number  after  3,  we  have  3  times  3,  5 
times  3,  7  times  3,  and  so  on ;  which  are  the  only  odd  numbers 
divisible  by  3. 

2d.  Taking  every  fifth  number  after  5,  we  have  3  times  5,  5 
times  5,  7  times  5,  and  so  on;  which  are  the  only  odd  numbers 
divisible  by  5.  And  the  same  will  be  true  of  every  other  number 
in  the  series.  Hence, 

3d.  If  we  cancel  every  third  number,  counting  from  3,  no 
number  divisible  by  3  will  be  left;  and  since  3  times  5  will  be 
canceled,  5  times  5  or  25,  will  be  the  least  composite  number  left 
in  the  series.  Hence, 

4th.  If  we  cancel  every  fifth  number,  counting  from  25,  no 
number  divisible  by  5  will  be  left;  and  since  3  times  7,  and  5 
times  7,  will  be  canceled,  7  times  7,  or  49,  will  be  the  least  com- 
posite number  left  in  the  series.  And  thus  with  all  the  prime 
numbers.  Hence, 

1  1O.    To  find  all  the  prime  numbers  within  any  given  limit, 

we  have  the  following 

* 
RULE.     I.    Write  all  the  odd  numbers  in  their  natural  order. 

II.  Cancel,  w  cross  out,  3  times  3,  or  9,  and  every  third  number 
after  it;  5  times  5,  or  25,  and  every  fifth  number  ifterit;  7  times 
7,  or  49,  and  every  seventh  number  after  it ;  and  so  on,  beginning 
with  the  second  power  of  each  prime  number  in  succession,  till  the 
given  limit  is  reached.  The  numbers  remaining,  together  with  the 
number  2,  will  be  the  prime  numbers  required. 

NOTBS. — 1.  It  is  unnecessary  to  count  for  every  ninth  number  after  9  times  9, 
for  being  divisible  by  3,  they  will  be  found  already  canceled;  the  same  may  be 
said  of  any  other  canceled,  or  composite  number. 

2.  This  method  of  obtaining  a  list  of  the  prime  numbers  was  employed  by 
Eratosthenes  (born  B.  C.,  275),  and  is  called  Eratosthenet'  Sieve. 


70  PROPERTIES  OF  NUMBERS. 

• 
TABLE  OF  PRIME  NUMBERS  LESS  THAN  1000. 


1 

59 

139 

233 

337 

439 

557 

653 

769 

883 

2 

61 

149 

239 

347 

443 

563 

659 

773 

887 

3 

67 

151 

241 

349 

449 

569 

661 

787 

907 

5 

71 

157 

251 

353 

457 

571 

673 

797 

911 

7 

73 

163 

257 

359 

461 

577 

677 

809 

919 

11 

79 

167 

263 

367 

463 

587 

683 

811 

929 

13 

83 

173 

269 

37o 

467 

593 

691 

821 

937 

17 

89 

179 

271 

379 

479 

599 

701 

823 

941 

19 

97 

181 

277 

383 

487 

601 

709 

827 

947 

23 

101 

191 

281 

389 

491 

607 

719 

829 

953 

29 

103 

193 

283 

397 

499 

613 

727 

839 

967 

31 

107 

197 

293 

401 

503 

617 

733 

853 

971 

37 

109 

199 

307 

409 

509 

619 

739 

857 

977 

41 

113 

211 

311 

419 

521 

631 

743 

859 

983 

43 

127 

223 

313 

421 

523 

641 

751 

863 

991 

47 

131 

227 

317 

431 

541 

643 

757 

877 

997 

53 

137 

229 

331 

433 

547 

647 

761 

881 

FACTORING. 
CASE    I. 

141.  To   resolve   any    composite    number  into   its 
prime  factors. 

The  Prime  Factors  of  a  number  are   those  prime  numbers 
which  multiplied  together  will  produce  the  given  number. 

142.  The  process  of  factoring  numbers  depends  upon  the  fol- 
lowing principles : 

I.  Every  prime  factor  of  a  number  is  an  exact  divisor  of  that 
number. 

.II.  The  only  exact  divisors  of  a  number  are  its  prime  factors,  or 
some  combination  of  its  prime  factors. 
1.  What  are  the  prime  factors  of  798  ? 

ANALYSIS.  Since  the  given  number  is  even,  we 
divide  by  2,  and  obtain  an  odd  number,  399,  for  a 
quotient  We  then  divide  by  the  prime  numbers 
3,  7,  and  19,  successively,  and  the  last  quotient  is 
1.  The  divisors,  2,  3,  7,  and  19,  are  the  prime 
factors  required,  (II).  Hence,  the 


OPERATION. 


2 
3 

7 
19 


79S 


399 
133 


19 


FACTORING.  71 

RULE.  Divide  the  given  number  l>y  any  prime  factor  ;  divide 
the  quotient  in  the  same  manner,  and  so  continue  the  division  until 
the  quotient  is  a  prime  number.  The  several  divisors  and  the  last 
quotient  will  be  the  prime  factors  required, 

PROOF.  The  product  of  all  the  prime  factors  wil1  be  the  given 
number. 

EXAMPLES    FOR   PRACTICE. 

1.  What  are  the  prime  factors  of  2150? 

2.  What  are  the  prime  factors  of  2445? 

3.  What  are  the  prime  factors  of  6300  ? 

4.  What  are  the  prime  factors  of  21504? 

5.  What  are  the  prime  factors  of  2366  ? 

6.  What  are  the  prime  factors  of  1000  ? 

7.  What  are  the  prime  factors  of  390625? 

8.  What  are  the  prime  factors  of  999999  ? 

143.  If  the  prime  factors  of  a  number  are  small,  as  2,  3,  5, 
7,  or  11,  they  may  be   easily  found  by  the  tests  of  divisibility, 
(13G),  or  by  trial.     But  numbers  may  be  proposed  requiring  many 
trials   to  find  their  prime   factors.       This  difficulty  is  obviated, 
within  a  certain  limit,  by  the  Factor  Table  given  on  pages  72,  73. 

By  prefixing  each  number  in  bold-face  type  in  the  column 
of  Numbers,  to  the  several  numbers  following  it  in  the  same  divis- 
ion of  the  column,  we  shall  form  all  the  composite  numbers  less 
than  10,000,  and  not  divisible  by  2,  3,  5,  7,  or  11;  the  numbers 
in  the  columns  of  Factors  are  the  least  prime  factors  of  the  num- 
bers thus  formed  respectively.  Thus,  in  one  of  the  columns  of 
Numbers  we  find  39,  in  bold-face  type,  and  below  39,  in  the  same 
column,  is  77,  which  annexed  to  39,  forms  3977,  a  composite  num- 
ber. The  least  prime  factor  of  this  number  is  41,  which  we  find 
at  the  right  of  77,  in  the  column  of  Factors. 

144,  Hence,  for  the  use  of  this  table,  we  have  the  following 
RULE.     I.    Cancel  from  the  given  number  all  factors  less  than 

13,  and  then  find  the  remaining  factors  by  the  table. 

II.  If  any  number  less  than  10,000  is  not  found  w  the  table, 
and  is  not  divisibk  by  2,  3,  5,  7,  or  11,  it  is  prime. 


72 


PROPERTIES  OF  NUMBERS. 


FACTOR  TABLE. 


1  < 

e  .. 

•2  o 

1  5 

1  I 

1  P 

j 

5    n 

•5  o 

1  t 

umbers, 
ictors. 

|j 

1  1 

1  « 

umbers, 
ictors. 

1 

99  29 

11  I7 

43  29 

79  37 

41  17 

83  17 

41  23 

09  31 

17  53 

77  31 

69  13 

9 

17  13 

57  19 

83  13 

51  13 

97  43 

59  17 

13  19 

27  29 

83  71 

2 

01  17 

57  31 

61  J7 

89  19 

77  13 

34 

69  53 

21  29 

47  47 

91  29 

21  13 

23  13 

69  13 

63  13 

91  47 

83  19 

01  19 

87  13 

31  61 

67  fff 

52 

47  13 

43  23 

15 

20 

25 

87  29 

03  41 

93  17 

43  43 

69  19 

07  41 

89  17 

49  13 

01  19 

21  4j 

01  41 

93  41 

19  ij 

39 

51  19 

71  13 

13  13 

99  13 

61  31 

13  17 

33  19 

07  23 

3O 

27  zj 

01  47 

69  17 

77  17 

19  17 

3 

89  23 

17  37 

41  13 

09  13 

07  31 

31  47 

37  31 

79  29 

48 

21  23 

23  17 

10 

37  29 

47  23 

33  17 

13  23 

39  19 

53  59 

81  13 

11  17 

39  13 

61  19 

03  17 

41  23 

59  29 

37  43 

29  13 

73  23 

59  37 

87  41 

19  61 

49  29 

77  13 

07  19 

77  19 

71  19 

61  13 

43  17 

81  59 

61  17 

93  23 

41  47 

51  59 

91  17 

27  13 

91  37 

77  31 

67  17 

53  43 

97  13 

73  29 

99<S3 

43  29 

63  19 

4 

37  17 

16 

21 

73  31 

71  37 

35 

77  41 

44 

47  37 

67  23 

03  13 

73  29 

33  23 

17  29 

81  29 

77  17 

03  31 

79  23 

27  19 

49  13 

87  17 

37  19 

79  13 

43  31 

19  13 

87  13 

97  19 

23  ij 

91  13 

2943 

53  23 

93  67 

81  13 

81  23 

49  17 

47  19 

99  23 

31 

51  5J 

40 

39  23 

5943 

53 

93  17 

11 

51  13 

59  17 

26 

03  29 

69  4j 

09  19 

53  61 

67  31 

11  47 

5 

21  19 

79  23 

71  i, 

03  19 

07  13 

87  17 

31  29 

69  41 

83  19 

17  13 

27  17 

39  17 

81  41 

73  41 

2343 

27  53 

89  J7 

33  37 

71  17 

91  67 

21  17 

29  23 

47  31 

91  19 

83  37 

27  37 

31  31 

99  59 

43  13 

89  67 

97  59 

29  73 

33  13 

57  13 

17 

97  13 

41  19 

33  13 

36 

61  31 

45 

49 

39  19 

51  19 

59  19 

03  13 

22 

69  17 

33  43 

01  ij 

63  17 

11  13 

01  13 

53  53 

59  13 

89  29 

11  29 

01  31 

27 

49  47 

11  ZJ 

69  13 

31  23 

13  17 

59  23 

89  19 

12 

17  17 

09  47 

01  37 

51  23 

29  19 

87  61 

37  13 

27  13 

63  31 

6 

07  17 

39  37 

27  17 

43  13 

61  zg 

49  41 

97  17 

41  19 

79  13 

714i 

11  13 

19  23 

51  17 

31  23 

47  41 

73  19 

53  73 

41 

53  29 

81  17 

77  19 

29  17 

41  17 

63  41 

49  13 

59  31 

93  31 

67  19 

17  23 

59  47 

97  19 

89  17 

67  23 

47  29 

69  29 

57  3'/ 

71  17 

97  23 

79  ij 

21  13 

73  17 

50 

54 

89  13 

61  13 

81  15 

63  31 

7347 

32 

83  29 

41  41 

77  23 

17  29 

29  61 

97  17 

71  31 

18 

79  43 

28 

11  13 

37 

63  23 

79  19 

2947 

47  13 

7 

73  19 

07  13 

91  29 

09  53 

33  53 

13  47 

71  43 

89  13 

41  71 

59  53 

03  19 

13 

17  23 

23 

13  29 

39  41 

21  61 

81  37 

46 

53  31 

61  43 

13  23 

13  13 

19  17 

23  23 

31  19 

47  17 

37  J7 

83  47 

01  43 

f.7  ij 

73  13 

31  17 

33  31 

29  31 

27  13 

39  17 

63  13 

43  19 

87  53 

07  17 

63  61 

91  17 

67  I  J 

39  13 

43  19 

29  17 

67  47 

77  29 

49  2j 

89  59 

19  31 

69  37 

97  23 

79  19 

43  17 

4943 

53  13 

69  19 

81  17 

57  ij 

99  13 

33  41 

83  13 

55 

93  13 

49  19 

53  17 

63  17 

73  13 

87  19 

63  53 

42 

61  59 

51 

la  37 

99  17 

57  23 

91  31 

69  23 

81  43 

93  37 

81  19 

23  41 

67  13 

11  19 

39  29 

8 

63  29 

19 

24 

99  13 

33 

91  17 

37  19 

81  31 

23  47 

43  23 

17  19 

69  37 

09  23 

07  29 

29 

17  ji 

99  29 

47  31 

87  43 

29  23 

49  JI 

41  29 

87  19 

19  19 

13  19 

11  41 

37  47 

38 

67  17 

93  13 

41  53 

61  67 

51  23 

91  13 

21  17 

19  41 

21  23 

41  ij 

09  ij 

43 

99  37 

43  J7 

67  19 

71  13 

14 

27  41 

49  31 

23  37 

49  17 

11  37 

03  13 

47 

49  19 

87  37 

93  19 

03  aj 

37  13 

61  23 

2929 

79  31 

27  43 

07  59 

09  17 

61  IJ 

97  29 

FACTORING. 


73 


FACTOR  TABLE  — CONTINUED. 


i 

i  s 

1| 

|  | 

lun'iberi. 
'actort. 

1  1 

<umbere. 
'actors. 

1! 

E  „• 

"s  1 

3     « 

1  \ 

ll 

°   rt 

111 

fc  £ 

*  ** 

56 

60 

39  47 

51  ij 

61  53 

57  IJ 

23  71 

13  47 

51  53 

53  19 

59  13 

03  ij 

01  17 

43  17 

59  19 

67  ij 

61  47 

27  2j 

17  19 

57  17 

5947 

71  19 

03  71 

19  ij 

63  zj 

77  ij 

77  19 

63  79 

33  Z9 

41  2J 

73  19 

63  59 

73  17 

11  ji 

23  19 

67  29 

87  71 

79  zg 

97  43 

47  ij 

5379 

79  13 

69  13 

83  23 

17  41 

31  J7 

87  ij 

89  8j 

89  J7 

77 

51  83 

71  43 

81  83 

71  73 

97 

27  17 

49  zj 

93  43 

93  61 

91  zj 

09  ij 

77  41 

73  37 

91  17 

87  37 

01  89 

29  ij 

59  73 

97  73 

69 

73 

29  59 

83  59 

79  61 

89 

99  17 

03  31 

33  4J 

71  ij 

99  67 

01  67 

03  67 

39  71 

81 

83  17 

03  29 

93 

07  17 

71  53 

77  jy 

65 

13  ji 

13  71 

47  61 

19  zj 

89  Ij 

09  59 

01  71 

27  71 

81  ij 

61 

09  zj 

29  i> 

19  ij 

51  zj 

31  47 

97  *9 

17  37 

07  41 

31  37 

99  41 

03  17 

11  17 

31  29 

27  17 

69  17 

37  79 

85 

27  79 

13  67 

61  43 

57 

07  Ji 

27  61 

43  5J 

39  41 

71  19 

43  17 

07  47 

47  aj 

22  19 

63  13 

07  ij 

09  41 

3347 

53  17 

61  17 

81  ji 

49  29 

0967 

57  ij 

47  ij 

73  29 

13  29 

19  Z9 

39  ij 

73  19 

63  37 

83  43 

53  ji 

31  19 

59  17 

53  47 

97  97 

23  59 

37  17 

41  ji 

89  Z9 

67  53 

87  ij 

59  41 

49  8j 

77  47 

67  17 

99  41 

29  17 

57  47 

57  79 

70 

73  73 

78 

77  ij 

51  17 

83  ij 

79  83 

98 

59  ij 

61  61 

83  zg 

03  47 

79  47 

01  29 

89  19 

57  43 

89  89 

89  41 

09  17 

67  7J 

69  ji 

93  19 

09  4J 

87  8j 

07  J7 

82 

67  ij 

93  17 

94 

27  31 

71  29 

79  J7 

66 

31  79 

91  19 

11  73 

01  59 

79  23 

90 

07  23 

41  13 

73  2j 

87  zj 

13  17 

33  ij 

97  IJ 

13  ij 

03  ij 

87  Ji 

17  71 

09  97 

4743 

77  53 

91  41 

17  ij 

37  Ji 

74 

31  41 

07  29 

93  ij 

19  29 

51  13 

5359 

58 

62 

23  J7 

61  zj 

09  ji 

37  17 

13  43 

86 

47  8j 

69  17 

69  71 

09  J7 

27  IJ 

31  19 

67  J7 

21  41 

49  47 

27  19 

11  79 

61  ij 

81  19 

81  41 

33  19 

33  zj 

4129 

81  73 

23  Ij 

59  29 

49  73 

21  J7 

71  47 

87  53 

93  13 

37  ij 

39  17 

47  17 

87  19 

29  17 

71  17 

51  J7 

33  89 

73  43 

95 

99  19 

9143 

41  79 

49  61 

93  41 

39  43 

91  ij 

57  zj 

39  Sj 

77  29 

03  13 

99 

93  71 

53  ij 

67  59 

97  47 

53  zg 

97  53 

79  vj 

51  41 

83  ji 

09  37 

13  23 

99  17 

83  61 

83  41 

99  ji 

63  17 

79 

99  4j 

53  17 

896i 

17  ji 

1747 

59 

89  19 

97  J7 

71 

71  ji 

13  41 

83 

71  ij 

91 

23  89 

37  19 

09  19 

63 

67 

11  ij 

93  59 

21  89 

03  19 

83  19 

01  19 

29  13 

43  61 

11  ZJ 

13  59 

07  19 

23  17 

75 

39  17 

21  5j 

87 

13  ij 

53  41 

53  37 

17  61 

19  71 

31  53 

41  J7 

01  ij 

43  ij 

33  13 

11  ji 

31  2J 

57  19 

59  23 

21  ji 

31  ij 

39  zj 

53  zj 

19  7j 

57  7j 

39  31 

17  2J 

39  ij 

63  73 

71  ij 

33  17 

41  17 

49  17 

57  17 

31  17 

61  19 

41  19 

49  ij 

434i 

71  17 

79  17 

41  ij 

71  zj 

51  4j 

63  ij 

43  19 

67  ji 

47  17 

59  19 

67  89 

77  61 

83  67 

47  19 

83  ij 

57  29 

69  67 

71  67 

69  ij 

57  61 

73  ji 

69  5j 

89  43 

91  97 

59  59 

64 

67  67 

71  71 

97  71 

79  79 

59  13 

77  67 

79  67 

93  53 

97  IJ 

63  67 

01  J7 

73  ij 

81  43 

76 

81  zj 

81  17 

91  59 

93  29 

99  29 

69  47 

03  19 

93  ij 

99  zj 

13  zj 

91  61 

83  83 

97  19 

97  17 

96 

77  43 

07  43 

68 

72 

19  19 

99  19 

99  37 

88 

92 

07  13 

83  ji 

09  ij 

17  17 

01  19 

27  zg 

80 

84 

01  ij 

11  61 

17  59 

89  5j 

31  59 

21  19 

23  31 

31  ij 

03  5j 

01  ji 

09  23 

17  13 

37  23 

93  13 

37  41 

47  41 

41  13 

33  17 

21  ij 

11  41 

43  37 

23  23 

41  31 

74  PROPERTIES  OF  NUMBERS. 

t 

1.  Resolve  1961  into  its  prime  factors. 

OPERATION.  ANALYSIS.     Cutting  off  the   two 

1961  —  37  =  53  rig^   hand    figures    of   the    given 

1961  =  37  X  53,  Ans.        number,  and  referring  to  the  table, 

column  No.,  we  find  the  other  part, 

19,  in  bold-face  type ;  and  under  it,  in  the  same  division  of  the  column, 
we  find  61,  the  figures  cut  off;  at  the  right  of  61,  in  column  Fac.,  we 
find  37,  the  lea'st  prime  factor  of  the  given  number.  Dividing  by  37, 
we  obtain  53,  the  other  factor. 

2.  Resolve  188139  into  its  prime  factors. 

OPERATION.  ANALYSIS.     "We  find    by   trial 


3 
7 

17 
17 


188139  *hat  the  given  number  is  divisible 


_.  by  3  and  7  ;  dividing  by  these  fac- 

— tors,  we  have  for  a  quotient  8959. 

8959  By  referring  to  the  factor  table, 

527  we  find  the  least  prime  factor  of 

this  number  to  be  17 ;  dividing  by 

17,  we   have  527  for  a  quotient, 


3x7x17x17x31,  Ans.      Referring  again  to  the  tatle,  we 

find  17  to  be  the  least  factor  of 
527,  and  the  other  factor,  31,  is  prime. 

EXAMPLES   FOR   PRACTICE. 

1.  Resolve  18902  into  its  prime  factors.       Ans.  2,  IS,  727. 

2.  Resolve  352002  into  its  prime  factors. 

3.  Resolve  6851  into  its  prime  factors. 

4.  Resolve  9367  into  its  prime  factors. 

5.  Resolve  203566  into  its  prime  factors. 

6.  Resolve  59843  into  its  prime  factors. 

7.  Resolve  9991  into  its  prime  factors. 

8.  Resolve  123015  into  its  prime  factors. 

9.  Resolve  893235  into  its  prime  factors. 

10.  Resolve  390976  into  its  prime  factors. 

11.  Resolve  225071  into  its  prime  factors. 

12.  Resolve  81770  into  its  prime  factors. 

13.  Resolve  6409  into  its  prime  factors. 

14.  Resolve  178296  into  its  prime  factors. 

15.  Resolve  714210  into  its  prime  factors. 


FACTORING.  75 


CASE  II. 

145.  To  find  all  the  exact  divisors  of  a  number. 

It  is  evident  that  all  the  prime  factors  of  a  number,  together 
with  all  the  possible  combinations  of  those  prime  factors,  will  con- 
stitute all  the  exact  divisors  of  that  number,  (142,  II). 

1 .  What  are  all  the  exact  divisors  of  360  ? 

OPERATION. 

360  =  1x2x2x2x3x3x5. 

1,2          4  ,       8     Combinations  of  1  and  2. 


Ans.< 


3  ,     6 

9  ,  18 

5  ,  10 

15  ,  30 

45  ,  90 


«  land  2  and  3. 


12  ,  24 
36  ,  72 
20  ,  40  "  "  "  1  and  2  and  5 

180      360  "  "land  2  and  3  and  5- 


ANALYSIS.  By  Case  I  we  find  the  prime  factors  of  360  to  be  1,  2. 
2,  2,  3,  3,  and  5.  As  2  occurs  three  times  as  a  factor,  the  different 
combinations  of  1  and  2  by  which  360  is  divisible  will  be  1,  1x2  =  2, 
1x2x2  =  4,  and  1x2x2x2  =  8;  these  we  write  in  the  first  line. 
Multiplying  the  first  line  by  3  and  writing  the  products  in  the  second 
line,  and  the  second  line  by  3,  writing  the  products  in  the  third  line, 
we  have  m  the  first,  second  and  third  lines  all  the  different  combina- 
tions of  1,  2,  and  3,  by  which  360  is  divisible.  Multiplying  the  first 
second  and  third  lines  by  5,  and  writing  the  products  in  the  fourth, 
fifth  and  sixth  lines,  respectively,  we  have  in  the  six  lines  together, 
every  combination  of  the  prime  factors  by  which  the  given  number^ 
360,  is  divisible. 

Hence  the  following 

RULE.     I.  Resolve  the  given  number  into  its  prime  factors. 

II  Form  a  series  having  1  for  the  first  term,  that  prime  factor 
which  occurs  the  greatest  number  of  times  in  the  given  number  for 
the  second  term,  the  square  of  this  factor  for  the  third  term,  and  so 
on,  till  a  term  is  reached  containing  this  factor  as  many  times  as  it 
occurs  in  the  given  number. 

III.  Multiply  the  numbers  in  this  line  by  another  factor,  jind 
these  results  by  the  same  factor,  and  so  on,  as  many  times  a*  thi* 
factor  occurs  in  the  given  number. 


76  PROPERTIES  OF  NUMBERS. 

IV.  Multiply  all  the  combinations  now  obtained  by  another 
factor  in  continued  multiplication,  and  thus  proceed  till  all  the  dif- 
ferent factors  have  been  used.  ALL  the  combinations  obtained  will 
be  the  exact  divisors  sought. 

EXAMPLES    FOR   PRACTICE. 

1.  What  are  all  the  exact  divisors  of  120  ? 

Ans.  1,2, 3, 4,  5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120. 

2.  Find  all  the  exact  divisors  of  84. 

Ans.  1,  2,  3,  4,  6,  7,  12,  14,  21,  28,  42,  84. 

3.  Find  all  the  exact  divisors  of  100. 

Ans.  1,  2,  4,  5,  10,  20,  25,  50,  100. 

4.  Find  all  the  exact  divisors  of  420. 

1,  2,  3,  4,  5,  6,  7,  10,  12,  14,  15,  20,  21,  28, 
.  30,  35,  42,  60,  70,  84,  105,  140,  210,  420. 

5.  Find  all  the  exact  divisors  of  1050. 

1,  2,  3,  5,  6,  7,  10,  14,  15,  21,  25,  30,  35,  42,  50,  70, 
75,  105,  150,  175,  210,  350,  525,  1050. 

GREATEST  COMMON  DIVISOR. 

94O.  A  Common  Divisor  of  two  or  more  numbers  is  a  number 
that  will  exactly  divide  each  of  them. 

147.  The  Greatest  Common  Divisor  of  two  or  more  numbers 
is  the  greatest  number  that  will  exactly  divide  each  of  them. 

148.  Numbers  Prime  to  each  other  are  such  as  have  no  com- 
mon divisor. 

NOTE. — A  common  divisor  is  sometimes  called  a  Common  Measure  ;  and  the 
greatest  common  divisor,  the  Greatest  Common  Measure. 

CASE    I. 

149.  When  the  numbers  can  be  readily  factored. 

It  is  evident  that  if  several  numbers  have  a  common  divisor, 
they  may  all  be  divided  by  any  component  factor  of  this  divisor, 
and  the  resulting  quotients  by  another  component  factor,  and  so 
on,  till  all  the  component  factors  have  been  used. 


GREATEST  COMMON  DIVISOR.  77 

I.  What  is  the  greatest  common  divisor  of  28,  140,  and  420  ? 

OPERATION.  ANALYSIS.     We  readily  see  that  7 

7    28  .  .  140  .  .  420          will  exactly  divide  each  of  the  given 
*     ~~7          M          ^          numbers ;    and  then,  4  will   exactly 
divide  each  of  the  resulting  quotients. 
*  '  '  '  Hence,  each   of  the   given  numbers 

4x7  =  28,  Ans.  can  he  exactly  divided  by  7  times  4 ; 
and  these  numbers  must  be  compo- 
nent factors  of  the  greatest  common  divisor.  Now,  if  there  were  any 
other  component  factor  of  the  greatest  common  divisor,  the  quotients, 
1,  5  and  15,  would  be  divisible  by  it.  But  these  quotients  are  prime 
to  each  other ;  therefore,  7  and  4  are  all  the  component  factors  of  the 
greatest  common  divisor  sought. 

From  this  analysis  we  derive  the  following 

RULE.     I.    Write  the  numbers  in  a  line,  with  a  vertical  line  at 
the  left,  and  divide  by  any  factor  common  to  all  the  numbers. 

II.  Divide  the  quotients  in  like  manner,  and  continue  the  divi- 
sion till  a  set  of  quotients  is  obtained  that  are  prime  to  each  other. 

II F.  Multiply  all  the  divisors  together,  and  the  product  will  be 
the  greatest  common  divisor  sought. 


EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  40,  75,  and  100  ? 

Ans.  5. 

2.  What  is  the  greatest  common  divisor  of  18,  30,  36,  42, 
and  54  ? 

3.  What  is  the  greatest  common  divisor  of  42,  63,  126,  and 
189?  Ans.  21. 

4.  What  is  the  greatest  common  divisor  of  135,  225,  270,  and 
315?  Ans.  45. 

5.  What  is  the  greatest  common  divisor  of  84,  126,  210,  252, 
294,  and  462  ? 

6.  What  is  the  greatest  common  divisor  of  216,  360,  432,  648, 
and  936  ?  Ans.  72. 

7.  What  is  the  greatest  common  divisor  of  102;  153,  and  255  ? 

Ans.  51. 
7* 


78  PROPERTIES  OF  NUMBERS. 

8.  What  is  the  greatest  common  divisor  of  756,  and  1575? 

9.  What  is  the  greatest  common  divisor  of  182, 864,  and  455? 

10.  What  is  the  greatest  common  divisor  of  2520,  and  3240  ? 

Am.  360. 

11.  What  is  the  greatest  common  divisor  of  1428,  and  1092  ? 

12.  What  is  the  greatest  common  divisor  of  1008,  and  1036? 

Ans.  28. 

CASE   II. 

IoO.  When  the  numbers  cannot  be  readily  factored. 
The  analysis   of  the  method  in  this   case   depends  upon  the 
following  properties  of  divisors. 

I.  An  exact  divisor  divides  any  number  of  times  its  dividend. 

II.  A  common  divisor  of  two  numbers  is  an  exact  divisor  of 
their  sum. 

III.  A  common  divisor  of  two  numbers  is  an  exact  divisor  of 
their  difference. 

NOTE. — The  last  two  properties  are  essentially  the  same  as  102,  II,  III. 

1.  What  is  the  greatest  common  divisor  of  527,  and  1207  ? 
OPERATION.  ANALYSIS.     We  will  first  describe  the  pro- 

1207          cess,  and  then  examine  the  reasons  for  the 


527 
459 


68 
68 


1054  several  steps  in  the  operation.  Drawing  two 
vertical  lines,  we  place  the  greater  number 
15o  on  the  right,  and  the  less  number  on  the  left, 
13(3  one  line  lower  down.  We  then  divide  1207, 
-TI  the  greater  number,  by  527,  the  less,  and 
v.  rite  the  quotient,  2,  between  the  verticals, 
the  product,  1054,  opposite  the  less  number  and  under  the  greater, 
and  the  remainder,  153,  below.  We  next  divide  527  by  this  re- 
mainder, writing  the  quotient,  3,  between  the  verticals,  the  product, 
459,  on  the  left,  and  the  remainder,  68,  below.  We  again  divide  the 
last  divisor,  153,  by  68,  and  obtain  2  for  a  quotient,  136  for  a  pro- 
duct, and  17  for  a  remainder,  all  of  which  we  write  in  the  same  order 
as  in  the  former  steps.  Finally,  dividing  the  last  divisor,  68,  by  the 
last  remainder,  17,  we  have  no  remainder,  and  17,  the  last  divisor,  is 
the  greatest  common  divisor  of  the  given  numbers. 

Now,  observing  that  the  dividend  is  always  the  sum  of  the  product 
and  remainder,  and  that  the  remainder  is  always  the  difference  of  the 


GREATEST  COMMON  DIVISOR. 


79 


OPERATION. 


527 

459 

68 
fW 

,  1 

,  1207 
1054 
153 
136 
17 

3 

2 

4 

dividend  and  product,  we  first  trace  the  work  in  the  reverse  order,  as 
indicated  by  the  arrow  line  in  the  diagram  below. 

17  divides  68,  as  proved  by  the 
last  division  ;  it  will  also  divide 
2  times  68,  or  136,  (I).  Now,  as 
17  divides  both  itself  and  136,  it 
will  divide  153,  their  sum,  (II). 
It  will  also  divide  3  times  153,  or 
459,  (I) ;  and  since  it  is  a  com- 
mon divisor  of  459  and  68,  it 
must  divide  their  sum,  527,  which 
is  one  of  the  given  numbers.  It 
will  also  divide  2  times  527,  or 
1054,  (I) ;  and  since  it  is  a  common  divisor  of  1054  and  153,  it  must 
divide  their  sum,  1207,  the  greater  number,  (II).  Hence,  17  is  a  com- 
mon divisor  of  the  given  numbers. 

Again,  tracing  the  work  in  the  direct  order,  as  indicated  in  the  fol- 
lowing diagram,  we  know  that 
the  greatest  common  divisor,  what- 
ever it  be,  must  divide  2  times 
527,  or  1054,  (I).  And  since  it 
will  divide  both  1054  and  1207, 
it  must  divide  their  difference, 
153,  (III).  It  will  also  divide  3 
times  153,  or  459,  (I) ;  and  as  it 

68     I — ? ,       136        will  divide  both  459  and  527,  it 

must  divide  their  difference,  68, 
(HI)-  It  will  also  divide  2  times 
68,  or  136,  (I) ;  and  as  it  will 

divide  both  136  and  153,  it  must  divide  their  difference,  17,  (III)  ; 
hence,  it  cannot  be  greater  than  17. 

Thus  we  have  shown, 

1st.  That  17  is  a  common  divisor  of  the  given  numbers. 
2d.  That  their  greatest  common  divisor,  whatever  it  be,  cannot 
be  greater  than  17.     Hence  it  must  be  17. 

From  this  example  and  analysis,  we  derive  the  following 

RULE.     I.  Draw  two  verticals,  and  write  the  two  numbers,  one 
on  each  side,  the  greater  number  one  line  above  the  less. 


527 


459 


1207 


1054 


153 


80 


PROPERTIES  OF  NUMBERS. 


II.  Divide  the  greater  number  by  the  less,  writing  the  quotient 
between  the  verticals,  the  product  under  the  dividend,  and  the  re- 
mainder below. 

III.  Divide  the  less  number  by  the  remainder ',  the  last  divisor 
by  the  last  remainder,  and  so  on,  till  nothing  remains.      The  last 
divisor  will  be  the  greatest  common  divisor  sought. 

IV.  If  more  than  two  numbers  be  given,  first  find  the  greatest 
common  divisor  of  two  of  them,  and  then  of  this  divisor  and  one 
of  the  remaining  numbers,  and  so  on  to  the  last  j  the  last  common 
divisor  found  will  be  the  greatest  common  divisor  required. 

NOTES. — 1.  When  more  than  two  nnmbera  are  given,  it  is  better  to  begin  with 
the  least  two. 

2.  If  at  any  point  in  the  operation  a  prime  number  occur  as  a  remainder,  it 
must  be  a  common  divisor,  or  the  given  numbers  have  no  common  divisor. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  18607  and  417979? 

OPERATION. 


417979 

18607 

2 

37214 

, 

45839 

2 

37214 

17250 

2 

8625 

1357 

6          8142 

966 

2 

483 

391 

1 

391 

368 

4 

~92 

Ans.        23 

4 

92 

2.  What  is  the  greatest  common  divisor  of  10661  and  123037 

OPERATION. 

12303 

10661 

1 

10661 

9852 

6 

1642 

Prime       809 

A~~     1 

GREATEST  COMMON  DIVISOR.  81 

3.  What  is  the  greatest  common  divisor  of  336  and  812  ? 

Ans.   28. 

4.  What  is  the  greatest  common  divisor  of  407  and  1067  ? 

5.  What  is  the  greatest  common  divisor  of  825  and  1372  T 

6.  What  is  the  greatest  common  divisor  of  2041  and  8476  ? 

Ans.   13. 

7.  What  is  the  greatest  common  divisor  of  3281  and  10778  ? 

8.  Find  the  greatest  common  divisor  of  22579,  and  116939. 

9.  What  is  the  greatest  common  divisor  of  49373  and  147731  ? 

Ans.   97. 

10.  What  is  the   greatest  common   divisor  of  1005973   and 
4616175  ? 

11.  Find  the  greatest  common  divisor  of  292,  1022,  and  1095. 

Ans.  73. 

12.  What  is  the  greatest  common  divisor  of  4718,    6951,  and 
8876?  Ans.  7. 

13.  Find  the  greatest  common  divisor  of  141,  799,  and  940. 

14.  What  is  the  greatest  common  divisor  of  484391  and  684877  ? 

Ans.   701. 

15.  A  farmer  wishes  to  put  364  bushels  of  corn  and  455  bushels 
of  oats  into  the  least  number  of  bins  possible,  that  shall  contain 
the  same  number  of  bushels  without  mixing  the  two  kinds  of 
grain ;  what  number  of  bushels  must  each  bin  hold  ? 

Ans.   91. 

16.  A  gentleman  having  a  triangular  piece  of  land,  the  sides  of 
which  are  165  feet,  231  feet,  and  385  feet,  wishes  to  inclose  it 
with  a  fence  having  pannels  of  the   greatest  possible  uniform 
length;  what  will  be  the  length  of  each  pannel? 

17.  B  has  $620,  C  $1116,  and  D  $1488,  with  which  they  agree 
to  purchase  horses,  at  the  highest  price  per  head  that  will  allow 
each  man  to  invest  all  his  money  j  how  many  horses  can  each  man 
purchase  ?  Ans.  B  5,  C  9,  and  D  12. 

18.  How  many  rails  will  inclose  a  field  14599  feet  long  by 
10361  feet  wide,  provided  the  fence  is  straight,  and  7  rails  high, 
and  the  rails  of  equal  length,  and  the  longest  that  can  be  used  ? 

Ans.  26880. 


£2  PROPERTIES  OP  NUMBERS. 

LEAST  COMMON  MULTIPLE. 

1*51.    A  Multiple  is  a  number  exactly  divisible  by  a  given 
number  ;  thus,  20  is  a  multiple  of  4. 

NOTES.  —  1.   A  multiple  is  necessarily  composite;   a  divisor  may  be  either 
prime  or  composite. 

2.  A  number  is  a  divisor  of  all  its  multiples  and  a  multiple  of  all  its  divisors. 


A  Common  Multiple  is  a  number  exactly  divisible  by 
two  or  more  given  numbers  ;  thus,  20  is  a  common  multiple  of  2, 
4,  5,  and  10, 

153.  The  Least  Common  Multiple  of  two  or  more  numbers 
is  the  least  number  exactly  divisible  by  those  numbers  ;  thus,  24 
is  the  least  common  multiple  of  3,  4,  6,  and  8. 

1«54.  From  the  definition  it  is  evident  that  the  product  of  two 
or  more  numbers,  or  any  number  of  times  their  product,  must  be 
a  common  multiple  of  the  numbers.  Hence,  A  common  multiple 
of  two  or  more  numbers  may  be  found  by  multiplying  the  given 
numbers  together. 

lo«5.   To  find  the  least  common  multiple. 

FIRST    METHOD. 

From  the  relations  of  multiple  and  divisor  we  have  the  following 
properties  : 

I.  A  multiple  of  a  number  must  contain  all  the  prime  factors 
of  that  number. 

II.  A  common  multiple  of  two  or  more  numbers  must  contain 
all  the  prime  factors  of  each  of  those  numbers. 

III.  The  least  common  multiple  of  two  or  more  numbers  must 
contain  all  the  prime  factors  of  each  of  those  numbers,  and  no 
other  factors. 

1.  Find  the  least  common  multiple  of  63,  66,  and  78. 

OPERATION.  ANALYSIS.      The 

63  =  3  X  3  X     7  number  cannot  be  less 

66  =  2x3x11  than   78,    because    it 

78  =  2  X  3  X   13  must  contain  78  ;  and 

2x3x  13  X  11x3x7  =  18018  An*,      if  it  contains   78,  it 

must  contain  all   its 
prime  factors,  viz.  ;         2  X  3  X  13. 


LEAST  COMMON  MULTIPLE.  83 

We  here  have  all  the  prime  factors,  and  also  all  the  factors  of  66 
ixcept  11.  Annexing  11  to  the  series  of  factors, 

2  X  3  X  13  X  11, 

and  we  have  all  the  prime  factors  of  78  and  66,  and  also  all  the  fac- 
tors of  63  except  one  3,  and  7.  Annexing  3  and  7  to  the  series  of 

factors, 

2  x  3  X  13  X  11  X  3  X  7, 

and  we  have  all  the  prime  factors  of  each  of  the  given  numbers,  and 
no  others;  hence  the  product  of  this  series  of  factors  is  the  least 
common  multiple  of  the  given  numbers,  (III). 

From  this  example  and  analysis  we  deduce  the  following 
RULE.     I.  Resolve  the  given  numbers  into  their  prime  factors. 
II.  Multiply  together  all  the  prime  factors  of  the  largest  number, 
and  such  prime  factors  of  the  other  numbers  as  are  not  found  in 
the  largest  number,  and  their  product  will  be  the  least  common 
multiple 

NOTR.  —  When  a  prime  factor  is  repeated  in  any  of  the  given  numbers,  it 
must  be  taken  as  many  times  in  the  multiple,  as  the  greatest  number  of  times  it 
appears  in  any  of  the  given  numbers. 

EXAMPLES   FOR   PRACTICE. 

1.  Find  the  least  common  multiple  of  60,  84,  and  132. 

Ans.  4620. 

2.  Find  the  least  common  multiple  of  21,  30,  44,  and  126. 

Ans.  13,860. 

3.  Find  the  least  common  multiple  of  8,  12,  20,  and  30. 

4.  Find  the  least  common  multiple  of  16,  60,  140,  and  210. 

Ans.  1,680. 

5.  Find  the  least  common  multiple  of  7,  15,  21,  25,  and  35. 

6.  Find  the  least  common  multiple  of  14,  19,  38,  42,  and  57. 

Ans.  798. 

7.  Find  the  least  common  multiple  of  144,  240,  480,  960. 


SECOND    METHOD. 

156.   1.  What  is  the  least  common  multiple  of  4,  9,  12,  18, 

and  36  ? 


84 


PROPERTIES  OF  NUMBERS. 


2 
2 

3 
3 

OPERATION 

4  .  .  9  .  .  12  .  . 

18 

.36 

2  .  .  9 

.  .    6  .  . 

9 

.  18 

9 
3 

.  .    3  .  . 

9 

.    9 

3 

3 

2x2x3x3  =  36  Am. 


ANALYSIS.  Wefirstwrite 
the  given  numbers  in  a  se- 
ries with  a  vertical  line  at 
the  left.  Since  2  is  a  fac- 
tor of  some  of  the  given 
numbers,  it  must  be  a  factor 
of  the  least  common  mul- 
tiple sought,  (155,111).  Di- 


2,3 

5 

2  .  . 

6  .  . 

3 

.  15 

5 

viding  as  many  of  the  numbers  as  are  divisible  by  2,  we  write  the 
quotients,  and  the  undivided  number,  9,  in  a  line  underneath.  Now, 
since  some  of  the  numbers  in  the  second  line  contain  the  factor  2,  the 
least  common  multiple  must  contain  another  2,  and  we  again  divide 
by  2,  omitting  to  write  any  quotient  when  it  is  1.  We  next  divide 
by  3  for  a  like  reason,  and  still  again  by  3.  By  this  process  we  have 
transferred  all  the  factors  of  each  of  the  numbers  to  the  left  of  the 
vertical ;  and  their  product,  36,  must  be  the  least  common  multiple 
sought,  (155,  III). 

2.  What  is  the  least  common  multiple  of  20,  12,  15,  and  75? 

OPERATION.  ANALYSIS.      We     readily 

2  ,  5  J  20  .  .  12  .  .  15  .  .  75  see  that  2  and  5  are  among 

the  factors  of  the  given  num- 
bers, and  must  be  factors  of 
the  least  common  multiple  ; 
hence,  writing  2  and  5  at  the 
left,  we  divide  every  number 

that  is  divisible  by  either  of  these  factors  or  by  their  product ;  thus, 
we  divide  20  by  both  2  and  5  ;  12  by  2  ;  15  by  5  ;  and  75  by  5.  We 
next  divide  the  second  line  in  like  manner  by  2  and  3 ;  and  afterward 
the  third  line  by  5.  By  this  process  we  collect  the  factors  of  the 
given  numbers  into  groups ;  and  the  product  of  the  factors  at  the 
left  of  the  vertical  is  the  least  common  multiple  sought. 

3.  What  is  the  least  common  multiple  of  7, 10, 15, 42,  and  70? 

ANALYSIS.  In  this  operation 
we  omit  the  7  and  10,  because 
they  are  exactly  contained  in 
some  of  the  other  given  numbers  ; 
thus,  7  is  contained  in  42,  and  10 
in  70  ;  and  whatever  will  contain 

42  and  70  must  contain  7  and  10.  Hence  we  have  only  to  find  th« 
least  common  multiple  of  the  remaining  numbers,  15,  42,  and  70. 


2x5x2x3x5  =  300,  Ans. 


3,7 
2,5 

OPERATION. 

15  .  .  42 

.70 

5  .  . 

2 

.  10 

3x7x2x5  =  210,  Ans. 


LEAST  COMMON  MULTIPLE.  85 

From  these  examples  we  derive  the  following 

RULE,  I.  Write  the  numbers  in  a  line,  omitting  such  of  the 
smaller  numbers  as  are  factors  of  the  larger,  and  draw  a  vertical 
line  at  the  left. 

II.  Divide  by  any  prime  factor  or  factors  that  may  be  contained 
in  one  or  more  of  the  given  numbers,  and  write  the  quotients  and 
undivided  numbers  in  a  line  underneath,  omitting  the  Vs. 

III.  In  like  manner  divide  the  quotients  and  undivided  numbers, 
and  continue  the  process  till  all  the  factors  of  the  given  numbers 
have  been  transferred  to  the  left  of  the  vertical.      Then  multiply 
these  factors  together,  and  their  product  will  be  the  least  common 
multiple  required. 

NOTE.  —  We  may  use  a  composite  number  for  a  divisor,  when  it  is  contained 
in  all  the  given  numbers. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  least  common  multiple  of  15,  18,  21,  24,  35, 
36,  42,  50,  and  60  ?  Ans.  12600. 

2.  What  is  the  least  common  multiple  of  6,  8,  10,  15,  18,  20, 
and  24  ?  Ans.  360. 

3.  What  is  the  least  common  multiple  of  9,  15,  25,  35,  45,  and 
100?  Ans.  6300. 

4.  What  is  the  least  common  multiple  of  18,  27,  36,  and  40  ? 

5.  What  is  the  least  common  multiple  of  12,  26,  and  52  ? 

6.  What  is  the  least  common  multiple  of  32,  34,  and  36  ? 

Ans.  4896. 

7.  What  is  the  least  common  multiple  of  8, 12,  18,  24,  27,  and 
36? 

8.  What  is  the  least  common  multiple  of  22,  33,  44,  55,  and 
66? 

1  9.  What  is  the  least  common  multiple  of  64,  84,  96,  and  216  ? 
10.  If  A  can  build  14  rods  of  fence  in  a  day,  B  25  rods,  C  8 
rods,  and  D  20  rods,  what  is  the  least  number  of  rods  that  will 
furnish  a  number  of  whole  days'  work  to  either  one  of  the  four 
men?  Ans.  1400. 


86  PROPERTIES  OF  NUMBERS. 

11.  What  is  the  smallest  sum  of  money  for  which  I  can  pur- 
chase either  sheep  at  $4  per  head,  or  cows  at  $21,  or  oxen  at  $49, 
or  horses  at  $72  ?  Am.  $3528 

12.  A  can  dig  4  rods  of  ditch  in  a  day,  B  can  dig  8  rods,  and 
C  can  dig  6  rods ;  what  must  be  the  length  of  the  shortest  ditch, 
that  will  furnish  exact  days'  labor  either  for  each  working  alone 
or  for  all  working  together  ?  Am.  72  rods. 

13.  The  forward  wheel  of  a  carriage  was  11  feet  in  circumfer- 
ence, and  the  hind  wheel  15  feet;  a  rivet  in  the  tire  of  each  was 
up  when  the  carriage  started,  and  when  it  stopped  the  same  rivets 
were  up  together  for  the  575th  time ;  how  many  miles  had  the 
carriage  traveled,  allowing  5280  feet  to  the  mile  ? 

Am.  17  miles  5115  feet. 

CANCELLATION. 

137.  Cancellation  is  the  process  of  rejecting  equal  factors 
from  numbers  sustaining  to  each  other  the  relation  of  dividend 
and  divisor. 

138.  It  is  evident  that  factors  common  to  the  dividend  and 
divisor  may  be  rejected  without   changing  the  quotient,  (117, 
III). 

1.  Divide  1365  by  105. 

OPERATION.  ^  ANALYSIS.     We  first  in- 

1QA_          ,        ,        ,,       1Q  dicate  the  division  by  wri- 

_  =  *  *—  =  13  ting  the  dividend  above  a 

105  £  X  £  X  $  horizontal  line  and  the  di- 

visor below.  Then  factor- 
ing each  term,  we  find  that  3,  5,  and  7  are  common  factors ;  and 
crossing,  or  canceling  these  factors,  we  have  13,  the  remaining  factor 
of  the  dividend,  for  a  quotient. 

139.  If  the  product  of  several  numbers  is  to  be  divided  by 
the  product  of  several  other  numbers,  the  common  factors  should 
be   canceled  before  the  multiplications   are   performed,  for  two 
reasons : 

1st.  The  operations  in  multiplication  and  division  will  thus  be 
abridged. 


CANCELLATION.  87 

2d.  The  factors  of  small  numbers  are  generally  more  readily 
detected  than  those  of  large  numbers. 

2.  Divide  20  times  56  by  7  times  15. 

OPERATION.  ANALYSIS.     Having  first  indi- 

4        g  cated  all  the  operations  required 

#0  X  fc$       32  by  the   question,  we   cancel  7 

~~7 77  =    „"  ==^*  from  7  and  56,  and  5  from  15 

and  20,  leaving  the  factors  3  in 
the  divisor,  and  8  and  4  in  the 

dividend.     Then  8  X  4  =  32,  which  divided  by  3,  gives  lOf ,  the  quo- 
tient required.      Hence  the  following 

RULE  I.  Write  the  numbers  composing  the  dividend  above  a 
horizontal  line,  and  the  numbers  composing  the  divisor  below  it. 

II.  Cancel  all  the  factors  common  to  both  dividend  and  divisor. 

III.  Divide  the  product  of  the  remaining  factors  of  the  dividend 
by  the  product  of  the  remaining  factors  of  the  divisor,  and  the 
result  will  be  the  quotient. 

NOTES.  —  1.  When  a  factor  is  canceled,  the  unit,  1,  is  supposed  to  take  its 
place. 

2.  By  many  it  is  thought  more  convenient  to  write  the  factors  of  the  dividend 
on  the  right  of  a  vertical  line,  aud  the  factors  of  the  divisor  on  the  left. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  quotient  of!8x6x4x42  divided  by  4  x  9 
X  3  x  7  x  6? 

FIRST   OPERATION.  SECOND    OPERATION. 

*£$  x  $  x  4  x  4&*       _  A  £ 


=  4 


4,  Ans. 

2.  Divide  the  product  of  21  x  8  X  60  x  8  x  6  by  7  X  12  X  3 
X  8  x  3.  Ans.  80. 

3.  The  product  of  the  numbers  16,  5,  14,  40;  16,  60,  and  50, 
is  to  be  divided  by  the  product  of  the  numbers  40,  24,  50,  20,  7, 
and  10;  what  is  the  quotient?  Ans.  32. 


88  PROPERTIES  OF  NUMBERS. 

4.  Divide  the  continued  product  of  12,  5,  183,  18,  and  70  by 
the  continued  product  of  3,  14,  9,  5,  20,  and  6. 

5.  If  213  x  84  x  190  x  264  be  divided  by  30  X  56  x  36, 
what  will  be  the  quotient  ? 

6.  Multiply  64  by  7  times  31    and  divide  the  product  by  8 
times  56,  multiply  this  quotient  by  15  times  88  and  divide  the 
product  by  55,  multiply  this  quotient  by  13  and  divide  the  pro- 
duct by  4  times  6.  Ans.  403. 

7.  How  many  cords  of  wood  at  $4  a  cord,  must  be  given  for  3 
tons  of  hay  at  $12  a  ton  ? 

8.  How  many  firkins  of  butter,  each  containing  56  pounds,  at 
15  cents  a  pound,  must  be  given  for  8  barrels  of  sugar,  each  con- 
taining 195  pounds,  at  7  cents  a  pound?  Ans.  13. 

9  A  grocer  sold  16  boxes  of  soap,  each  containing  66  pounds 
at  9  cents  a  pound,  and  received  as  pay  99  barrels  of  potatoes, 
each  containing  3  bushels ;  how  much  were  the  potatoes  worth  a 
bushel  ? 

10.  A  farmer  exchanged  480  bushels  of  corn  worth  70  cents  a 
bushel,  for  an  equal  number  of  bushels  of  barley  worth  84  cents  a 
bushel,  and  oats  worth  56  cents  a  bushel ;  how  many  bushels  of 
each  did  he  receive  ?  Ans.  240. 

11.  A  merchant  sold  to  a  farmer  two  kinds  of  cloth,  one  kind  at 
75  cents  a  yard,  and  the  other  at  90  cents,  selling  him  twice  as 
many  yards  of  the  first  kind  as  of  the  second.    He  received  as  pay 
132  pounds  oi  butter  at  20  cents  a  pound ;   how  many  yards  of 
each  kind  of  cloth  did  he  sell  ? 

Ans.  22  yards  of  the  first,  and  11  yards  of  the  second. 

12.  A  man  took  six  loads  of  potatoes  to  market,  each  load  con- 
taining 20  bags,  and  each  bag  2  bushels.     He  sold  them  at  44 
cents  a  bushel,  and  received  in  payment  8  chests  of  tea,  each  con- 
taining 22  pounds  j  how  much  was  the  tea  worth  a  pound  ? 

Ans.  60  cents. 


DEFINITIONS,  NOTATION,  AND  NUMERATION.  89 

FRACTIONS. 

DEFINITIONS,  NOTATION,  AND  NUMERATION. 

160.  When  it  is  necessary  to  express  a  quantity  less  than  a 
unit,  we  may  regard  the  unit  as  divided  into  some  number  of  equal 
parts,  and  use  one  of  these  parts  as  a  new  unit  of  less  value  than 
the  unit  divided.     Thus,  if  a  yard,  considered  as  an  integral  unit, 
be  divided  into  4  equal  parts,  then  one,  two,  or  three  of  these 
parts  will  constitute  a  number  less  than  a  unit.     The  parts  of  a 
unit  thus  used  are  called  fractional  units  ;  and  the  numbers  formed 
from  them,  fractional  numbers.     Hence 

161.  A  Fractional  unit  is  one  of  the  equal  parts  of  an  inte- 
gral unit. 

162.  A  Fraction  is  a  fractional  unit,  or  a  collection  of  frac- 
tional units. 

163.  Fractional  units  take  their  name,  and  their  value,  from 
the  number  of  parts  into  which  the  integral  unit  is  divided.    Thus, 

If  a  unit  be  divided  into  2  equal  parts,  one  of  the  parts  is 
called  one  half.  If  a  unit  be  divided  into  3  equal  parts,  one  of  the 
parts  is  called  one  third.  If  a  unit  be  divided  into  4  equal  parts, 
one  of  the  parts  is  called  one  fourth. 

And  it  is  evident  that  one  third  is  less  in  value  than  one  half, 
one  fourth  less  than  one  third,  and  so  on. 

164.  To  express  a  fraction  by  figures,  two  integers  are  re- 
quired ;  one  to  denote  the  number  of  parts  into  which  the  inte- 
gral unit  is  divided,  the  other  to  denote  the  number  of  parts  taken, 
or  the  number  of  fractional  units  in  the  collection.     The  former 
is  written  below  a  horizontal  line,  the  latter  above.     Thus, 

One  half  is  written         A          One  fifth  is  written         | 


One  third         " 


Two  thirds       «  | 

One  fourth       "  \ 
Two  fourths     " 

Three  fourths  "  | 


Two  fifths          "  | 

One  seventh      "  ^ 

Three  eighths  "  f 

Five  ninths       "  $ 


Eight  tenths 
8* 


90  FRACTIONS. 

165.  The  Denominator  of  a  fraction  is  the  number  below  the 
line. 

It  denominates  or  names  the  fractional  unit,  and  it  shows  how 
many  fractional  units  are  equal  to  an  integral  unit. 

166.  The  Numerator  is  the  numbei  above  the  line. 

It  numerates  or  numbers  the  fractional  units;  and  it  shows 
how  many  are  taken. 

167.  The  Terms  of  a  fraction  are  the  numerator  and  deno- 
minator, taken  together. 

168.  Since  the  denominator  of  a  fraction  shows  how  many 
fractional  units  in  the  numerator  are  equal  to  1  integral  unit,  it 
follows, 

I.  That  the  value  of  a  fraction  in  integral  units,  is  the  quo- 
tient of  the  numerator  divided  by  the  denominator. 

II.  That   fractions    indicate  division,  the    numerator  being  a 
dividend  and  the  denominator  a  divisor. 

169.  To  analyze  a  fraction  is  to  designate  and  describe  its 
numerator  and  denominator.     Thus  |  is  analyzed  as  follows  :  — 

7  is  the  denominator,  and  shows  that  the  units  expressed  by  the 
numerator  are  sevenths. 

5  is  the  numerator,  and  shows  that  5  sevenths  are  taken. 

5  and  7  are  the  terms  of  the  fraction  considered  as  an  expres- 
sion of  division,  5  being  the  dividend  and  7  the  divisor. 

EXAMPLES   FOR   PRACTICE. 

Express  the  following  fractions  by  figures  :  — 

1.  Four  ninths.  Ans.  £. 

2.  Seven  Jiffy-sixths.  Ans.  ^. 

3.  Sixteen  forty-eighths. 

4.  Ninety-five  one  hundred  seventy-ninths. 

5.  Five  hundred  thirty-six  four  hundredths. 

6.  One  thousand  eight  hundred  fifty-seven  nine  thousand  Jive 
hundred  twenty-firsts. 

7.  Twenty-five  thousand  eighty-sevenths. 

8.  Thirty  ten  thousand  eighty-seconds. 

9.  One  hundred  one  ten  millionths. 


DEFINITIONS,  NOTATION,  AND  NUMERATION.  91 

Read  and  analyze  the  following  fractions  :  — 

10-        7  Tft8«;  *iii;  M- 


n.  V;  «5  T¥O;  fl;  4%;  'I5;  W;  £%• 
12.  Ill;  HI;  'W°5  TB%O;  188?- 

1  Q   150.  436.  13785.  150072.  100001 
id'  537  >  97^;  47955?  475000  >  2000U2' 

Fractions  are  distinguished  as  Proper  and  Improper. 

170.  A  Proper  Fraction  is  one  whose  numerator  is  less  than 
its  denominator;  its  value  is  less  than  the  unit  1.  i 

171.  An  Improper  Fraction  is  one  whose  numerator  equals 
or  exceeds  its  denominator;  its  value  is  never  less  than  the  unit  1. 

NOTES.  —  1.  The  value  of  a  proper  fraction,  always  being  less  than  a  unit,  can 
only  be  expressed  in  a  fractional  form  ,  hence,  its  name. 

2.  The  value  of  an  improper  fraction,  always  being  equal  to,  or  greater  than 
a  unit,  can  always  be  expressed  in  some  other  form;  hence  its  name. 

172.  A  Mixed  Number  is  a  number  expressed  by  an  integer 
and  a  fraction. 

173.  Since  fractions  indicate  division,  (1G8,  II),  all  changes 
in  the  terms  of  a  fraction  will  affect  the  value  of  that  fraction  ac- 
cording to  the  laws  of  division;  and  we  have  only  to  modify  the 
language  of  the  General  Principles  of  Division,  by  substituting 
the  words  numerator,  denominator,  and  fraction,  or  value  of  the 
fraction,  for  the  words  dividend,  divisor,  and  quotient,  respectively, 
and  we  shall  have  the  following 

GENERAL   PRINCIPLES    OF   FRACTIONS. 

174.  PRIN.    I.       Multiplying    the   numerator    multiplies    the 
fraction,  and  dividing  the  numerator  divides  the  fraction. 

PRIN.  II.  Multiplying  the  denominator  divides  the  fraction, 
and  dividing  the  denominator  multiplies  the  fraction. 

PRIN.  III.  Multiplying  or  dividing  both  terms  of  the  fraction 
by  the  same  number,  does  not  alter  the  value  of  the  fraction. 

1  7«5,   These  three  principles  may  be  embraced  in  one 

GENERAL   LAW. 

A  change  in  the  NUMERATOR  produces  a  LIKE  change  in  tht 
value  of  the  fraction  ;  but  a  change  in  the  DENOMINATOR  produce* 
an  OPPOSITE  change  in  the  value  of  the  fraction. 


92  FRACTIONS. 

REDUCTION. 

1  76.  The  Reduction  of  a  fraction  is  the  process  of  changing 
its  terms,  or  its  form,  without  altering  its  value. 

CASE   I. 

177.   To  reduce  fractions  to  their  lowest  terms. 
A  fraction  is  in  its  lowest  terms  when  its  numerator  and  denomi- 
nator are  prime  to  each  other;  that  is,  when  both  terms  have  no 
common  divisor. 

1.  Reduce  the  fraction  -f^  to  its  lowest  terms. 

OPERATION.  ANALYSIS.    Dividing  both  terms  of 

_6  o    __  1  2  __  4  the  fraction  by  the  same  number  does 

~Qr  not  alter  the  value  of  the  fraction, 

15)  -60    -L  4  (174,   HI);    hence,    we   divide   both 

terms  of  ^  by  5,  and  both  terms  of 

the  result,  £f  ,  by  3,  and  obtain  4  for  the  final  result.     As  4  and  7  are 
prime  to  each  other,  the  lowest  terms  of  T6ff°-  are  $. 

Instead  of  dividing  by  the  factors  5  and  3  successively,  we  may 
divide  by  the  greatest  common  divisor  of  the  given  terms,  and  reduce 
the  fraction  to  its  lowest  terms  at  a  single  operation.  Hence,  the 

RULE.  Cancel  or  reject  aU  factors  common  to  both  numerator 
and  denominator.  Or, 

Divide  both  terms  by  their  greatest  common  divisor. 

EXAMPLES    FOR    PRACTICE. 

1.  Reduce  T7223  to  its  lowest  terms.  Ana.  f 

2.  Reduce  f  J  to  its  lowest  terms.  Ans.  J. 

3.  Reduce  |||  to  its  lowest  terms.  Ans.  f  . 


4.  Reduce  T7g55  to  its  lowest  terms.  Ans.  §. 

5.  Reduce  ^J|  to  its  lowest  terms.  Ans.  |. 

6.  Reduce  |||  to  its  lowest  terms. 

7.  Reduce  |||  to  its  lowest  terms. 

8.  Reduce   69   to  its  lowest  terms. 


9.  Reduce  f  §f  f  to  its  lowest  terms. 

10.  Reduce  f  £?  to  its  lowest  terms.  Ans. 
NOTE.  —  Consult  the  factor  table. 

11.  Reduce  T825T^  to  its  lowest  terms.  Ans. 


.REDUCTION.  93 

12.  Reduce  |J|J  to  its  lowest  terms.  Ans.   ff . 

13.  Reduce  f  if  f  to  its  lowest  terms. 

14.  Reduce  f  |]|  to  its  lowest  terms.  Ans.   Ji. 

15.  Reduce  |||jj,  and  |£jff  to  their  lowest  terms. 

CASE    II. 

178.  To  reduce  an  improper  fraction  to  a  whole  or 
mixed  number. 

1.  Reduce  2T937  to  a  whole  or  mixed  number. 

OPERATION. 

2Ty  =  297  -*- 12  =  24T%  =  24|. 

ANALYSIS.  Since  the  value  of  a  fraction  in  integral  units  is  equal 
to  the  quotient  of  the  numerator  divided  by  the  denominator,  (168,  1,) 
we  divide  the  given  numerator,  297,  by  the  given  denominator,  12, 
and  obtain  for  the  value  of  the  fraction,  the  mixed  number  24/2  =  ^4f . 
Hence  the 

RULE.      Divide  the  numerator  by  the  denominator. 

NOTKS.  1.  When  the  denominator  is  an  exact  divisor  of  the  numerator,  the 
result  will  be  a  whole  number. 

2.  In   all  answers  containing  fractions,  reduce   the  fractions  to  their  lowest 
terms. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  9f  to  an  equivalent  integer.  Ans.  16, 

2.  Reduce  636  to  an  equivalent  integer. 

3.  Reduce  *|4  to  a  mixed  number.  Ans.  17J. 

4.  Reduce  3T439  to  a  mixed  number.  Ans.  26||. 

5.  Reduce  5r^f  to  a  mixed  number,  Ans.  24|. 

6.  Reduce  9^/  to  a  mixed  number.  Ans.  17^|. 

7.  Reduce  2||l  to  a  mixed  number. 

8.  Reduce  'f  J|5  to  a  mixed  number.  Ans.  156|. 

9.  Reduce  3||2  to  a  mixed  number. 

10.  Reduce  3/f/  to  a  mixed  number.  Ans.  4|J. 

11.  Reduce  2|f  |5  to  a  mixed  number.  Ans.  100|. 

12.  Reduce  {{JJgf  to  a  mixed  number. 

13.  In  78g59  of  a  day  how  many  days?         Ans.  982|  days. 

14.  In  4T°/  of  a  dollar  how  many  dollars?  Ans.  $31  A. 

15.  If  1000  dollars  be  distributed  equally  among  36  men,  \vhat 
part  of  a  dollar  must  each  man  receive  in  change  ?         Ans.   J. 


94  FRACTIONS. 

CASE  in. 

179.  To  reduce  a  whole  number  to  a  fraction  having 
a  given  denominator. 

1.   Reduce  37  to  an  equivalent  fraction  whose  denominator  shall 

be  5. 

OPERATION.  ANALYSES.    Since  in  each  unit  there  are 

37  v  5  =  185  5  fifths,  in  37  units  there  must  be  37  times 

37  ==  i|s,  Ans.          5  fifth8,  or  185  fifths  =  ]f5,     The  nume- 
rator, 185,  is  obtained  in  the  operation  by 

multiplying  the  whole  number,  37,  by  the  given  denominator,  5. 
Hence  the 

RULE.  Multiply  the  whole  number  by  the  given  denominator ; 
take  the  product  for  a  numerator,  under  which  write  the  given  de- 
nominator. 

NOTE. — A  whole  number  may  be  reduced  to  a  fractional  form  by  writing  1 

9 

r 


under  it  for  a  denominator ;  thus,  °  =  9 


EXAMPLES    FOR    PRACTICE. 

1.  Reduce  17   to  an    equivalent   fraction  whose    denominator 
shall  be  6.  Ans.   Jg2. 

2.  Change  375  to  a  fraction  whose  denominator  shall  be  8. 

3.  Change  478  to  a  fraction  whose  denominator  shall  be  24. 

4.  Reduce  36  pounds  to  ninths  of  a  pound. 

5.  Reduce  359  days  to  sevenths  of  a  day.  Ann.  25713. 

6.  Reduce  763  feet  to  fourteenths  of  a  foot.        Ans.   J  °T6?82. 

7.  Reduce  937  to  a  fractional  form.  Ans.   9p. 

CASE   IV. 

18O.   To  reduce  a  mixed  number  to  an  improper  frac- 
tion. 

1.  In  12^  how  many  sevenths? 

OPERATION.  ANALYSIS.     In  the  whole  number  12,  there  are 

12f  12  X  7  sevenths  =  84  sevenths,  (Case  III),  and 

7  84  sevenths  -f  5  sevenths  =-.  89  sevenths,  or  8,p. 

89  Hence  the  following 


REDUCTION.  95 

RULE.  Multiply  the  whole  number  by  the  denominator  of  the 
fraction;  to  the  product  add  the  numerator,  and  under  the  sum 
write  the  denominator. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  15|  to  fifths.  Ans.  \9. 

2.  Reduce  24|  to  an  improper  fraction.  Ans.   9^9. 

3.  Reduce  57f  to  an  improper  fraction. 

4.  Reduce  356 jf  to  an  improper  fraction.  Ans.   6ff 4. 

5.  Reduce  872T52  to  an  improper  fraction. 

6.  Reduce  300^^  to  an  improper  fraction.  Ans.   9^J§1. 

7.  Reduce  434Jf  to  an  improper  fraction.  Ans.   103°300. 

8.  In  15|  how  many  eighths  ? 

9.  In  135^  how  many  twentieths?  Am.  2-J$3. 

10.  In  43|  bushels  how  many  fourths  of  a  bushel  ? 

11.  In  760T9<j  days  how  many  tenths  of  a  day? 

CASE  v. 
181.    To  reduce  a  fraction  to  a  given  denominator. 

We  have  seen  that  fractions  may  be  reduced  to  lower  terms  by 
division.  Conversely, 

I.  Fractions  may  be  reduced  to  higher  terms  by  multiplication. 

II.  All   higher  terms  of  a  fraction  must  be  multiples  of  its 
lowest  terms. 

1.  Reduce  |  to  a  fraction  whose  denominator  is  40. 

o 

OPERATION,  ANALYSIS.     We  first  divide  40,  the  re- 

40  -r-  8  =  5  quired  denominator,  by  8,  the  denomi- 

3x5  nator  of  the  given  fraction,  to  ascertain 

8  X   5  =  ^'  Am         if  ii;  be  a  multiPle  of  this  term>  8-     The 
division  shows  that  it  is  a  multiple,  and 

that  5  is  the  factor  which  must  be  employed  to  produce  it.  We  there- 
fore multiply  both  terms  of  £  by  5,  (174,  III),  and  obtain  |§,  the  re- 
quired result.  Hence  the 

RULE  Divide  the  required  denominator  by  the  denominator 
of  the  given  fraction,  and  multiply  both  terms  of  the  fraction  by 
the  quotient. 


9(J  FRACTIONS. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  |  to  a  fraction  having  24  for  a  denominator. 

Ans.  Jf 

2.  Reduce  •£%  to  a  fraction  whose  denominator  is  96. 

Ans.  f|. 

3.  Reduce  ||  to  a  fraction  whose  denominator  is  51. 

4.  Reduce  Ts^  to  a  fraction  whose  denominator  is  78. 

5.  Reduce  -ff-g  to  a  fraction  whose  denominator  is  3000. 

Ans.  ^V 

6.  Change  7|  to  a  fraction  whose  denominator  is  8. 

7.  Change  16375  to  a  fraction  whose  denominator  is  176. 

8.  Change  5T3T  to  a  fraction  whose  denominator  is  363- 

9.  Change  36|  to  a  fraction  whose  denominator  is  42. 

Ans.   'fl2. 

CASE    VI. 

183.  To  reduce  two  or  more  fractions  to  a  common 
denominator. 

A  Common  Denominator  is  a  denominator  common  to  two  or 
more  fractions. 

1.  Reduce  |  and  ^  to  a  common  denominator. 

ANALYSIS.     We  multiply  the  terms  of  the 

OPERATION.  first  fraction  by  the  denominator  of  the  second, 

3  X   9  _  27  an(j  the  terms  of  the  second  fraction  by  the 

5x9  denominator  of   the  first,    (174,    III).     This 

7x5  must  reduce  each  fraction  to  the  same  deno- 

q         r  ===  4  5  minator,  for  each  new  denominator  will  be  the 

product  of  the  given  denominators.   Hence  the 

RULE.     Multiply  the  terms  of  each  fraction  ly  the  denominators 
of  all  the  other  fractions 
NOTE.— Mixed  numbers  must  first  be  reduced  to  improper  fractions. 

EXAMPLES    FOR    PRACTICE. 

1.  Reduce  |  and  |  to  a  common  denominator.     Ans.  jf ,  T^. 
2    Reduce  §  and  $  tc  a  common  denominator. 


REDUCTION.  97 

3.  Reduce  f  ,  T52  and  ^    to  a  common  denominator. 

A™-  j&,  M,  Mr 

4.  Reduce  J,  5|  and  If  to  equivalent  fractions  having  a  com- 

mon denominator.  Ans.  ||,  L7^0,  ||. 

5.  Reduce  T45  and  T37  to  a  common  denominator. 


6.  Reduce  ^,  ^  and  T1T  to  a  common  denominator. 

7.  Reduce  ^,  |,  T72  and  T9ff  to  a  common  denominator. 


CASE   VII. 

183.  To  reduce  fractions  to  their  least  common  de- 
nominator. 

The  Least  Common  Denominator  of  two  or  more  fractions  is 
the  least  denominator  to  which  they  can  all  be  reduced. 

184.  We  have  seen  that  all  higher  terms  of  a  fraction  must 
be  multiples  of  its  lowest  terms,  (181,  II).     Hence, 

I.  If  two  or  more  fractions  be  reduced  to  a  common  denomi- 
nator, this  common  denominator  will  be  a  common  multiple  of  the 
several  denominators. 

II.  The  least  common  denominator  must  therefore  be  the  least 
common  multiple  of  the  several  denominators. 

1.  Reduce  |,  ,72  and  T\  to  their  least  common  denominator. 

OPERATION.  ANALYSIS.     We  first  find  the  least 


3   ,   5 


12   . .   15  common  multiple  of  the  given  deno- 


„        ^  minators,  which  is  60.     This  must 

be  the  least  common  denominator  to 


3x5x2x2  =  60          which  the  given  fractions  can  be  re- 

duced,  (II).  Reducing  each  frac- 
tion  to  the  denominator  60,  by  Case 
V,  we  obtain  |g,  jjj|  and  B8S  for  the 
answer.  Hence  the  following 

RULE.     I.  Find  the  hast  common  multiple  of  the  given  denom- 
j  for  the  least  common  denominator. 
9  G 


98  FRACTIONS. 

II.  Divide  this  common  denominator  by  each  of  the  given  de- 
nominators, and  multiply  each  numerator  l>y  the  corresponding 
quotient.  The  products  will  be  the  new  numerators. 

NOTES.  —  1.  If  the  several  fractions  are  not  in  their  lowest  terms,  they  should 
be  reduced  to  their  lowest  terms  before  applying  the  rule. 

2.  When  two  or  more  fractions  are  reduced  to  their  least  common  denominator, 
their  numerators  and  the  common  denominator  will  be  prime  to  each  other. 

EXAMPLES   FOR   PRACTICE. 

1.  Eeduce  f  and  r^  to  their  least  common  denominator. 

Ans.  JJ,  if. 

2.  Reduce  |,  |  and  |  to  their  least  common  denominator. 

3.  Reduce  |,  -^  and  |J  to  their  least  common  denominator. 

4.  Reduce  |,  |  and  |  to  their  least  common  denominator. 

5.  Reduce  3^,  ^J  and  ||  to  their  least  common  denominator. 


6.  Reduce  |,  T^>  ||  and  ^  to  their  least  common  denominator. 

Atts     52     24     75       8 

^.ns.  ?g,  ^g,  ^g,  ^5. 

7.  Reduce  2|,  T7^,  ^  and  |J  to  their  least  common  denomi- 
nator. Ans.  |I§,  T^,  ^  T%. 

8.  Reduce  |^,  59g  and  |J  to  their  least  common  denominator. 

9.  Reduce  |§,  T\5^  and  £|  to  their  least  common  denominator 

^»«-  88,  88,  f  I- 

10.  Reduce  2^,  -j7^  and  ^||  to  their  least  common  denomi- 
nator. Ans.  2VT,  Jjf,  2620T- 

11.  Reduce  J|J,  ||f  and  j|||  to  their  least  common  denomi- 
nator. Ans.  ^Vs,  &V*,  iVA- 

12.  Reduce  2f  ,  ,\  and  1^  to  their  least  common  denominator. 

13.  Reduce  T9BV^,  f^  and  |J||  to  their  least  common  denom 
inator.  /n..  ifjfi|,  JHf?  j,  i||H|. 

14.  Reduce  ^,  jl,  T25,  287,  -g93  and  |J  to  their  least  common 
denominator.  J^.  f  |gg,  «f  |g,  ;jg§,  §|Jg,  *f  jj,  «§i8- 

15.  Reduce  4,  r^,  rj\,  573  and   ^  to  their  least  common  de- 
nominator. 

16.  Reduce  T\,  7\,  ||  and  4^  to  their  least  common   denomi- 
nator. Ans.  TVfc,  T  jj,  T%,  $$J. 


ADDITION.  99 

ADDITION. 

185.  The  denominator  of  a  fraction  determines  the  value  of 
the  fractional  unit,  (1G5);  hence, 

I.  If  two  or  more  fractions  have  the  same  denominator,  their 
numerators  express  fractional  units  of  the  same  value. 

II.  If  two  or  more  fractions  have  diiferent  denominators,  their 
numerators  express  fractional  units  of  different  values. 

And  since  units  of  the  same  value  only  can  be  united  into  one 
sum,  it  follows, 

III.  That  fractions  can  be  added  only  when  they  have  a  com- 
mon denominator. 

1.  What  is  the  sum  of  \,  T63  and  T\  ? 

OPERATION. 

'   1J       6  •>  12  +  25  +  8 

i  +  A  +  ts  = go =  n  =  f 

ANALYSIS.  "We  first  reduce  the  given  fractions  to  a  common  deno- 
minator, (III).  And  as  the  resulting  fractions,  -' §,  f  £ ,  and  ^  have  the 
same  fractional  unit,  (I),  we  add  them  by  uniting  their  numerators 
into  one  sura,  making  ^  —  £,  the  answer. 

2.  Add  5|,  3J  and  4T73. 

ANALYSIS.      The   sum   of   the 

OPERATION.  integers,  5,  3,  and  4,  is  12;  the 

'    74  =       2  s  sum  of  the  fractions,  -J,  },  and 

4  +  H  +  12  ~     ~J-&  T-L,  is  2/5 .     Hence,  the  sum  of 

142\,  Ans.      both   fractions    and    integers   is 

12  +  2/3  =  14&. 

18G.  From  these  principles  and  illustrations  we  derive  the 
following  general 

RULE.  I.  To  add  fractions. —  When  necessary,  reduce  the.  frac- 
tions to  their  least  common  denominator ;  then  add  the  numerators 
and  place  the  sum  over  the  common  denominator. 

II.  To  add  mixed  numbers.  —  Add  the  integers  and  fractions 
separately,  and  then  add  their  sums. 

NOTE. — All  fractional  results  should  be  reduced  to  their  lowest  terms,  and  if 
improper  fractions,  to  whole  or  mixed  numbert. 


FRACTIONS. 


EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  sum  of  T72,  T42,  ft  and  jj  ?  ^4ws.  2J. 

2.  What  is  the  sum  of  if,  ft,  ft  and  T83  ?  ^Ina.  1  j. 

3.  What  is  the  sum  of  /T,  281?  if  and  £f  ? 

4.  What  is  the  sum  of  7j  J,  8§|,  2||,  51  1  and  4||  ? 

^Iws.  28|. 

5.  What  is  the  sum  of  37&,  12f  J,  13f  J  and  ff  ? 

6.  Add  I,  |  and  |. 

7.  Add  |,  I,  £f  and  ft.  J.TIS.  2^. 

8.  Add  |,  §  and  ft. 

9.  Add  ft,  if  and  &.  ^  1£  j. 

10.  Add  |,  I  IJ  and  f  J.  ^7is.  2|9. 

11.  Add  |,  i«,  ||,  If  and  |f.  ^«s  4ftV 

12.  Add  3£,  4|  and  2ft. 

13.  Add  16ft  and  24ft.  Ant.  40^. 

14.  Add  1^,  2|,  3j,  4|  and  5|. 

15.  Add  4ft,  836T  and  2&.  Ans.  14j|. 

16.  Add  1,  1,  ft  and  ft.  Ans.  ff 

17.  Add  J,  |,  ft  and  ft. 

18.  Add  J,  I,  T3T,  ft  and  ^|.  ^ns.  Iff. 

19.  Add  i,  ft,  ft  and  | 

20.  Add  411,  105|,  300{,  241|  and  472f        ^w».  1161|J. 

21.  Add  4|,  21,  1ft,  2&,  5ft,  7|,  4^  and  6|. 

22.  Four  cheeses  weighed  respectively  36|,  42|,  39T\  and  51| 
pounds;  what  was  their  entire  weight?         Ans.  169||  pounds. 

23.  What  number  is  that  from  which  if  4|  be  taken,  the  re- 
mainder will  be  3||?  Ans.  8|. 

24.  What  fraction  is  that  which  exceeds  T55  by  557  ? 

25.  A  beggar  obtained  ^  of  a  dollar  from  one  person,  J  from 
another,  -^  from  another,  and  ft  from  another  ;  how  much  did  he 
get  from  all  ? 

26.  A  merchant  sold  46  j  yards  of  cloth  for  $127  ft,  64^  yards 
for  $226|,  and  76|  yards  for  $31  2|  ;  how  many  yards  of  cloth  did 
he  sell,  and  how  much  did  he  receive  for  the  whole  ? 

Ans.  l87|j  yards,  for  $666{g. 


SUBTRACTION.  101 

SUBTRACTION. 

187.    The  process  of  subtracting  one  fraction  from  another  is 
based  upon  the  following  principles : 

I.  One  number  can  be  subtracted  from  another  only  when  the 
two  numbers  have  the  same  unit  value.     Hence, 

II.  In  subtraction  of  fractions,  the  minuend  and  subtrahend 
must  have  a  common  denominator,  (1853  I). 

1.  From  |  subtract  |. 

OPERATION.  ANALYSIS.      Reducing   the 

4 2  __  1 2-  \  o  __  _2  given  fractions  to  a  common 

denominator,     the    resulting 

fractions  jf  and  }§  express  fractional  units  of  the  same  value,  (185, 
I).  Then  12  fifteenths  less  10  fifteenths  equals  2  fifteenths  =  T2?,  the 
answer. 

2.  From  238J  take  24|. 

OPERATION.  ANALYSIS.     We  first  reduce  the  frac- 

2331  _  238-3.  tional  parts,  \  and  £,  to  the  common 

945  24i°  denominator,    12.      Since   we   cannot 

take  ]f  from  ^,  we  add  1  =  jf,  to  T\, 
213T52  Am.        making  i| .     Then,  |j|  subtracted  from 
-f  4  leaves  fa ;  and  carrying  1  to  24,  the  integral  part  of  the  subtrahend, 
(73,  II),  and  subtracting,  we  have  213/2  f°r  tne  entire  remainder. 

188.  From  these  principles  and  illustrations  we  derive  the 
following  general 

RULE.  I.  To  subtract  fractions. —  When  necessary,  reduce  the 
fractions  to  their  least  common  denominator.  Subtract  the  nume- 
rator of  the  subtrahend,  from  the  numerator  of  the  minuend,  and 
place  the  difference  of  the  new  numerators  over  the  common  denom- 
inator. 

II.  To  subtract  mixed  numbers.  —  Reduce,  the  fractional  parts 
to  a  common  denominator,  and  then  subtract  the  fractional  and 
integral  parts  separately. 

NOTK. — "We  may  reduce  mixed  numbers  to  improper  fractions,  and  subtract 
by  the  rule  for  fractions.     But  this  method  generally  imposes  the  useless  labor 
of  reducing  integral  numbers  to  fractions,  and  fractions  to  integers  again. 
Q* 


102  FRACTIONS. 

EXAMPLES    FOR   PRACTICE. 

1.  From  T73  take  T3^.  Am.  T4^. 

2.  From  ||  take  f  J.  Ans.  |. 

3.  From  ff  take  /3. 

4.  From  §  take  |.  <4ns.  ^\. 
5  From  |  take  fa 

6.  From  if  take  fa  JLws.  f  £. 

7  From  T\  take  £f .  ^TW.  /F. 

8.  From  |*  take  £f .  .4ns.  J^. 

9.  From  /T  take  y1^, 

10.  From  T72  take  fe  Ans.  £f . 

11.  From  g\  take  y1^.  ^ns.  ^. 

12.  From  3%  take  ^ 

13.  From  16|  take  7J-  ^TIS.  9f . 

14.  From  36£J  take  8^|.  ^Ins.  27|. 

15.  From  25T70  subtract  14{f. 

16.  From  75  subtract  4|.  Ans.  70|. 

17.  From  18|  subtract  5|. 

18.  From  265\  subtract  25{|. 

19.  From  28^f  subtract  3T\.  ^jw.  24j^. 

20.  From  78/5  subtract  32|. 

21.  The  sum  of  two  numbers  is  26^,  and  the  less  is  7^  ;  what 
is  the  greater?  Ans.  19^j. 

22.  What  number  is  that  to  which  if  you  add  18^,  the  sum 
will  be  97}  ? 

23.  What  number  must  you  add  to  the  sum  of  126|  and  240 f, 
to  make  560f  ?  Ans.  193|». 

24.  What  number  is  that  which,  added  to  the  sum  of  £,  y1^, 
and  T!H,  will  make  ||  ?  Ans.  fa 

25.  To  what  fraction  must  |  be  added,  that  the  sum  may  be  |  ? 
26    From  a  barrel  of  vinegar  containing  31  £  gallons,  14i  gallons 

were  drawn  ;  how  much  was  then  left?  AUK.  16f  gallons. 

27.  Bought  a  quantity  of  coal  for  $140f,  and  of  lumber  for 
1456$ .  Sold  the  coal  for  $775£,  and  the  lumber  for  $516T3g  ;  how 
much  was  my  whole  gain  ?  Ans.  $694f|. 


THEORY  OF  MULTIPLICATION  AND  DIVISION.  1Q3 


THEORY  OF  MULTIPLICATION  AND  DIVISION  OF  FRACTIONS. 

189.  In  multiplication  and  division  of  fractions,  the  various 
operations  may  be  considered  in  two  classes : 

1st.  Multiplying  or  dividing  a  fraction. 
2d.  Multiplying  or  dividing  by  a  fraction. 

190.  The  methods  of  multiplying  and  dividing  fractions  may 
be  derived  directly  from   the  General  Principles  of  Fractions, 
(174);  as  follows: 

I.  To  multiply  a  fraction. — Multiply  its  numerator  or  divide  its 
denominator,  (174,  I.  and  IT). 

II.  To  divide  a  fraction. — Divide  its  numerator  or  multiply  its 
denominator,  (174,  I.  and  II). 

GENERAL    LAW. 

III.  Perform  the  required  operation  upon  the  numerator,  or  the 
opposite  upon  the  denominator ,  (174,  III). 

191.  The  methods  of  multiplying  and  dividing  by  a  fraction 
may  be  deduced  as  follows : 

1st.  The  value  of  a  fraction  is  the  quotient  of  the  numerator 
divided  by  the  denominator  (1O8,  I)-  Hence, 

2d.  The  numerator  alone  is  as  many  times  the  value  of  the 
fraction,  as  there  are  units  in  the  denominator. 

3d.  If,  therefore,  in  multiplying  by  a  fraction,  we  multiply  by 
the  numerator,  this  result  will  be  too  great,  and  must  be  divided 
by  the  denominator. 

4th.  But  if  in  dividing  by  a  fraction,  we  divide  by  the  nume- 
rator, the  resulting  quotient  will  be  too  small,  and  must  be  multi- 
plied by  the  denominator. 

Hence,  the  methods  of  multiplying  and  dividing  by  a  fraction 
may  be  stated  as  follows : 

I.  To  multiply  by  a  fraction.  —  Multiply  by  the  numerator  and 
divide  by  the  denominator,  (3d). 

II.  To  divide  by  a  fraction. — Divide  by  the  numerator  and  mul- 
tiply by  the  denominator,  (4th). 


104  FRACTIONS. 

GENERAL   LAW. 

III.  Perform  the  required  operation  by  the  numerator  and  the 
opposite  by  the  denominator. 

MULTIPLICATION. 
1.  Multiply  T%  by  4. 

FIRST  OPERATION.  ANALYSIS.     In  the  first  opera- 

T52  x  4  =  f  i  =  If  tion,  we  multiply  the  fraction 

by  4  by  multiplying  its  nume- 

SECOND    OPERATION.  ^    by  4.    and    jn    ^    gec()nd 

T2  X  4  —  3  =  lg  operation,  we  multiply  the  frac- 

THIRD    OPERATION.  tl0n  ^  4  ^  divi^S  it8  d^0m' 

5         v  inator  by  4,  (190,  I  or  III). 

T—  X  ^  =  -|  ==  If  In  the  third  operation,  we  ex- 


press the  multiplier  in  the  form 


of  a  fraction,  indicate  the  mul- 
tiplication, and  obtain  the  result  by  cancellation. 

2.  Multiply  21  by  f 

FIRST  OPERATION.  ANALYSIS.     To  multiply  by  4, 

2ix4_=84._i2  we  must  multiply  by  4  and  di- 

vide by  7,  (191,  I  or  III). 

SECOND  OPERATION.  In  the  first  operation,  we  first 

21x4  =  3x4  =  12  multiply  21  by  4,  and  then  di- 

vide the  product,  84,  by  7. 

THIRD  OPERATION.  In  the  second  operation,  we 

3  first  divide  21  by  7,  and  then 

^A  x  ^  __  ^2  multiply  the  quotient,  3,  by  4. 

1        f  In  the  third  operation,  we  ex- 

press the  whole  number,  21,  in 

the  form  of  a  fraction,  indicate  the  multiplication,  and  obtain  the 
result  by  cancellation. 

3.  Multiply  T\  by  |. 

FIRST  OPERATION.  ANALYSIS.     To  multiply  by 

l«t  step,     «,  x  7  =  |f  !»  we  must  multiply  by  7  and 

M,UP,   fJ~8=//3  divide  by  8,  (Ml,  I  or  III). 

In  the  first  operation,  we  muL 
TT3Z  =  TB    An8-       tiply  A  by  7  and  obtain  f  |  ; 


MULTIPLICATION.  105 

SECOND  OPERATION.  we  then  divide  ?|  by  8  and  obtain 

r$j  x  |  =  T3T52  =  T5g  fVV  which  reduced   gives   T5ff,  the 

required   product.     In   the   second 

operation  we  obtain  the  same  result 

_  x    —=  J>  by  multiplying  the  numerators  to- 

AP       8  gether  for  the  numerator  of  the  pro- 

duct, and  the  denominators  together 

for  the  denominator  of  the  product.    In  the  third  operation,  we  indicate 
the  multiplication,  and  obtain  the  result  by  cancellation. 

193*  From  these  principles  and  illustrations  we  derive  the 
following  general 

RULE.  I.  Reduce  all  integers  and  mixed  numbers  to  improper 
fractions. 

II.  Multiply  together  the  numerators  for  a  new  numerator,  and 
the  denominators  for  a  new  denominator. 

NOTES. — 1.  Cancel  all  factors  common  to  numerators  and  denominators. 
2.  If  a  fraction  be  multiplied  by  its   denominator,  the  product  will  be  the 
numerator. 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  I  by  8.  Am.  2| 

2.  Multiply  *  by  27,  ^  by  4,  and  &  by  9. 

3.  Multiply  ^  by  15.  Ans.  §. 

4.  Multiply  8  by  |.  Ans.  6. 

5.  Multiply  75  by  T3g,  7  by  38T,  756  by  f,  and   572  by  2\. 

6.  Multiply  f  by  f . 

7.  Multiply  Ji  by  ||,  and  If  by  f  J. 

8.  Multiply  /2  by  |J,  and  ft  by  §J. 

9.  Multiply  24  by  3j.  Ans.  10. 

10.  Multiply  If  by  lj|.  ^ns.   2J. 

11.  Multiply  A  by  2i|. 

12.  Find  the  value  of  f  X  |  X  ^  X  J.  ^*s.  5\. 

13.  Find  the  value  of  f  X  J  X  Jf  X  T4T  X  4f      -4«s.  T15. 

14.  Find  the  value  of  ||  X  57^  X  jf  |. 

15.  Find  the  value  of  2|  x  2^  x  r\  X  TJH  X  1T75  X  26J. 

^«s.  2. 

16.  Find  the  value  of  T7T  x  25T  X  4|  x  15  x  T3ff. 

17.  Find  the  value  of    *&  x  5^  X  %°.  -4n«.  T^. 


106  FRACTIONS. 


18.  Find  the  value  of  (4£  X  J)  +  If  X  (3}  —  $j). 

19.  Find  the  value  of  28  +  (7|  —  2|)  x  §  X  (f  +  -J). 

NOTB  3.  —  The  word  of  between  fractions  is  equivalent  to  the  sign  of  multi- 
plication;   and   such   an   expression   is     sometimes  called  a  compound  fraction. 

Find  the  values  of  the  following  indicated  products  :  — 

20.  I  of  <  of  f.  An*.  f  . 

21.  §  of  I  of  3\.  An*.   rV 

22.  f  of  A  of  f 

23.  }i  of  A  of  if  of  fi.  An,.  ft. 

24.  £  of  |  of  |  of  I  of  §  of  *  of  I  of  |  of  T9j. 

In  the  following  examples,  cancellation  may  be  employed  by  the 
aid  of  the  Factor  Table. 

25.  What  is  the  value  of  f  £  J  x  |f  jf  X  ||||  ?      Jr?s.  T%3T. 

26.  What  is  the  value  of  §f  jj  x  f  f  jj  |  X  j  j§J  ?      4n«.  §4. 

27.  What  is  the  value  of  {}}f  x  ||f|  X  |f  |f  ? 


28.  What  will  7  cords  of  wood  cost  at  $3f  per  cord? 

^s.  $25|. 

29.  What  is  the  value  of  (|)2  X  If  X  (f)6?      Ans.   2T|7^. 

30.  If  1  horse  eat  |  of  a  bushel  of  oats  in>a  day,  how  many 
bushels  will  10  horses  eat  in  6  days  ?  Ans.  25|. 

31.  What  is  the  cube  of  12J  ? 

32.  At  $9|  per  ton,  what  will  be  the  cost  of  £  of  |  of  a  ton  of 
hay?  Ans.  $4. 

33.  At  $T9ff  a  bushel,  what  will  be  the  cost  of  If  bushels  of 
corn  ? 

34.  When    peaches  are  worth  $J  per  basket,  what  is  ^  of  a 
basket  worth  ? 

35.  A  man  owning  |  of  156|  acres  of  land,  sold  -^  of  |  of  his 
share;  how  many  acres  did  he  sell?  Ans.  47. 

36.  What  is  the  product  of  (|)s  x  (^)3  X  (A)1  X  (3j)*? 

Ans.  fjf. 

37.  If  a  family  consume  1£  barrels  of  flour  a  month,  how  many 
barrels  will  6  families  consume  in  8T9^  months? 

38.  What  is  the  product  of  150}—  (j  of  121|+jof  48|)—  75^ 
multiplied  by  3  x  (|  of  1J  X  4)  —  2J  ?  Ans.  342  T<\^. 


DIVISION.  107 

39.  A  man  at  his  death  left  his  wife  $12,500,  which  was  £  of 
|   of  his  estate;  she   at  her  death   left  |  of  her  share  to  her 
daughter ;  what  part  of  the  father's  estate  did  the  daughter  re- 
ceive ?  Ans.  || . 

40.  A  owned  |  of  a  cotton  factory,  and  sold  |  of  his  share  to 
B,  who  sold  ^  of  what  he  bought  to  C,  who  sold  |  of  what  he 
bought  to  D ;  what  part  of  the  whole  factory  did  each  then  own  ? 

Ans.  A,  fa  ;  B,  T% ;  C,  /g  ;  D,  -fa. 

1  41.  What  is  the  value  of  2jx~£-f£  of  4^  x  (|)2+(3§)8— (3f)«? 

Ans.  36Jfj. 
DIVISION. 

194.    1.  Divide  |J  by  3. 

ANALYSIS.      In  the  first  ope- 

FIRST    OPERATION.  ration  we  diyide  the    fraction  by 

|i  -^-  3  =  T7T  3  by  dividing  its  numerator  by 

SECOND  OPERATION.  3'  ***  .in  the  second  °Peration 

2 !    .    q  2 1  _     7  we  d^v^e  t*16  fraction  by  3  by 

25~        ~"75==35  multiplying  its  denominator  by 

3,  (190,  II  or  III). 

2.  Divide  15  by  f 

ANALYSIS.    To  divide  by  5J-,  we 
FIRST  OPERATION.  ^  divide  by  3  and  multiply 

by  7,  (191,  II  or  III). 

SECOND  OPERATION.  In  the  first  operation,  we  first 

15  H-  |  =  105  ~  3  =  35         divide  15  by  3,  and  then  mul- 
tiply the  quotient  by  7. 

In  the  second  operation  we  first  multiply  15  by  7,  and  then  divide 
the  product  by  3. 

3.  Divide  T4,  by  f. 

FIRST  OPERATION.  ANALYSIS.     To  divide  by 

1st  step,  T<t  —  3  =  -^  §>  we  must  divide  by  3  and 

2d  step,  A  X  6  -  fl  -  I  Ans.        multiP^  bJ  5'  d91,  II  or 

III).     In  the  first  operation 

SECOND  OPERATION.  we  first  divide  T4j  by  3  by 

y*j  X  |  ==  JS  =  |  multiplying  the  denomina^ 

tor  by  3.    We  then  multi- 


108  FRACTIONS. 

THIRD  OPERATION.  ply  the  result,  j43,  by  5,  by 

4         k  multiplying  the  numerator 

£jj  x  3"  =  I  by  5,  giving  |  J  =  £  for  the 

required  quotient.      By  in- 
specting this  operation,  we 

observe  that  the  result,  f  §,  is  obtained  by  multiplying  the  denomi- 
nator of  the  given  dividend  by  the  numerator  of  the  divisor,  and  the 
numerator  of  the  dividend  by  the  denominator  of  the  divisor.  Hence, 
in  the  second  operation,  we  invert  the  terms  of  the  divisor,  f,  and 
then  multiply  the  upper  terms  together  for  a  numerator,  and  the 
lower  terms  together  for  a  denominator,  and  obtain  the  same  result  as 
in  the  first  operation.  In  the  third  operation,  we  shorten  the  pro- 
cess by  cancellation. 

We  have  learned  (1OT)  that  the  reciprocal  of  a  number  is  1 
divided  by  the  number.  If  we  divide  1  by  |,  we  shall  have  1  -7- 
I  =  1  X  I  -  |.  Hence 

195.   The  Reciprocal  of  a  Fraction  is  the  fraction  inverted. 

From  these  principles  and  illustrations  we  derive  the  following 
general 

RULE.  I.  Reduce  integers  and  mixed  numbers  to  improper 
fractions. 

II.   Multiply  the  dividend  by  the  reciprocal  of  the  divisor. 

Noras.  —  1.  If  the  vertical  line  be  used,  the  numerators  of  the  dividend  and 
the  denominators  of  the  divisor  must  be  written  on  the  right  of  the  vertical. 

2.  Since  a  compound  fraction  is  an  indicated  product  of  several  fractions,  its 
reciprocal  may  be  obtained  by  inverting  each  factor  of  the  compound  fraction. 

EXAMPLES   FOB   PRACTICE. 

1.  Divide  if  by  4.  if  X  i  -  A,  An,. 

2.  Divide  j?  by  5,  and  jff  by  80. 

3.  Divide  10  by  §.  Ans.  35. 

4.  Divide  28  by  |.  and  3  by  ,fi3. 

5.  Divide  56  by  If.  Ans.  36. 

6.  Divide  £J  by  f . 

7.  Divide  1$  by  j,  if  by  fa  and  3|  by  5f 

8.  Divide  1J  by  1£.  Ans.  If. 

9.  Divide  IJf  by  jj.  Ans.  If. 


DIVISION.  109 

10.  Divide  f  of  j  by  J  of  TV 

OPERATION.  ANALYSIS.    The  dividend, 

3  x  5  __  i  reduced  to  a  simple  fraction, 

7  x  _5    __  _5  is  £  ;   the   divisor,  reduced 

i  x   i  8  __  6  _  1  1  Ans                *n  **ke  manner,  is  T5B  ;  and 

*  i  divided  by  T%  is  14,  the 

'  quotient  required.     Or,  we 

.3    V    5    V    ^    V    *—  ^     _    1  JL 

'    &  &  may  apply  the  general  rule 

directly  by  inverting  both  factors  of  the  divisor,. 

NOTE  3.  —  The  second  method  of  solution  given  above  has  two  advantages. 
1st,  It  gives  the  answer  by  a  single  operation;  2d,  It  affords  greater  facility  for 
cancellation. 

11.  Divide  |  of  T5T  by  T8T  of  T%.  Ans.  1. 

12.  Divide  T72  of  T63  by  J  of  /T.  Ans.  l|i. 

13.  Divide  2£  x  7J  by  3^  x  3^. 

14.  Divide  11  by  |  X  5^  x  7. 

15.  Divide  3|  x  19  by  £  x  7|  X  If.  ^»s.  25. 

16.  Divide  Ts3  X  if  by  £  X  j'x  T57  X  ||  X  f^. 

Ans.   3  1  5. 

17.  Divide  f  f  £  by  -£%.  ^4»s.  12V 

18.  Divide  2%3fV  by  fj  X  ||  X  §4.  Ans.  tf. 

19.  Divide  i  X  i  X  |  X  f  by  f  *  f  X  f  X  I  X  j%. 

20.  What  is  the  value  of  j|  ? 

OPERATION. 


ANALYSIS.  The  fractional  form  indicates  division,  the  numerator 
being  the  dividend  and  the  denominator  the  divisor,  (168,  II)  ;  hence, 
•sve  reduce  the  mixed  numbers  to  improper  fractions,  and  then  treat 
the  denominator,  2-/,  as  a  divisor,  and  obtain  the  result,  1^,  by  the 
general  rule  for  division  of  fractions. 

NOTE  4.  —  Expressions  like  -^  and  -^  are  sometimes  called  complex  fractions. 
47  V 

5.  In  the  reduction  of  complex  fractions  to  simple  fractions,  if  either  the 
numerator  or  denominator  consists  of  one  or  more  parts  connected  by  +  or  —  , 
the  operations  indicated  by  these  signs  must  first  be  performed,  and  afterward 
the  division. 

21.  What  is  the  value  of  i?  Ans.   «. 


HO  FRACTIONS. 

22.  What  is  the  value  of  f    X  **  ?  4ns.  2. 

A  X  5i 

23.  What  is  the  value  of  ^iM  ? 


3 


_ 

24.  Reduce  j—    ^  to  its  simplest  form. 

3  -r  I 

5  _  2 

25.  Reduce  ?  -  1  to  its  simplest  form. 

i  X  f 

26.  Reduce  J  —  ^7-  to  its  simplest  form.  Ans.  ||. 

bs  —  °TS 

27.  If  7  pounds  of  coffee  cost  $|,  how  much  will  1  pound  cost  ? 

28.  If  a  boy  earn  $|  a  day,  how  many  days  will  it  take  him  to 
earn  $6^?  Ans.   17f 

29.  If  |  of  an  acre  of  land  sell  for  $30,  what  will  an  acre  sell 
for  at  the  same  rate  ?  Ans.   $67^. 

30.  At  ^  of  |  of  a  dollar  a  pint,  how  much  wine  can  be  bought 
for  $T%  ?  Ans.   2|  pints. 

31.  If  -j3^  of  a  barrel  of  flour  be  worth  $2£,  how  much  is  1 
barrel  worth  ?  Ans.   $7f  . 

32.  Bought  £  of  4^  cords  of  wood,  for  |  of  |  of  $30;  what 
was  1  cord  worth  at  the  same  rate?  Ans. 

33.  If  235£  acres  of  land  cost  $1725|,  how  much  will 
acres  cost?  Ans.  $918|4f. 

34.  Of  what  number  is  26|  the  |  part?  Ans.   31^. 

35.  The  product  of  two  numbers  is  27,  and  one  of  them  is  2|  ; 
what  is  the  other  ? 

36.  By  what  number  must  you  multiply  16]4  to  produce  148|  ? 

37.  What  number  is  that  which,  if  multiplied  by  |  of  |  of  2, 
will  produce  |  ?  Ans.   l|i. 

38.  Divide  720  —  (|  X  28^77^)  by  40^  -f  (^  H-  |)  X  Q)*. 

39.  What  is  the  value  of  (3£  X  (|)2  +  |  of  ^)3  -f-  (l7j  —  | 


.GREATEST  COMMON  DIVISOR. 


GREATEST  COMMON  DIVISOR  OF  FRACTIONS. 

196.  The  Greatest  Common  Divisor  of  two  or  more  fractions 
is  the  greatest  number  which  will  exactly  divide  each  of  them, 
giving  a  whole  number  for  a  quotient. 

NOTE.  —  The  definition  of  an  exact  divisor,  (128),  is  general,  and  applies  to 
fractions  as  well  as  to  integers. 

197.  In  the  division  of  one  fraction  by  another  the  quotient 
will  be  a  whole  number,  if,  when  the  divisor  is  inverted,  the  two 
lower  terms  may  both  be  canceled.     This  will  be  the  case  when 
the  numerator  of  the  divisor  is  exactly  contained  in  the  numerator 
of  the   dividend,  and    the   denominator  of  the   divisor    exactly 
contains,  or  is  a  multiple  of,  the  denominator  of  the  dividend. 
Hence, 

I.  A  fraction  is  an  exact  divisor  of  a  given  fraction  when,  its 
numerator  is  a  divisor  of  the  given  numerator,  and  its  denominator 
is  a  multiple  of  the  given  denominator.     And, 

II.  A  fraction  is  a  common  divisor  of  two  or  more  given  frac- 
tions when  its  numerator  is  a  common  divisor  of  the  given  nume- 
rators, and  its  denominator  is  a  common  multiple  of  the  given 
denominators.     Therefore, 

III.  The  greatest  common  divisor  of  two  or  more  given  frac- 
tions is  a  fraction  whose  numerator  is  the  greatest  common  divisor 
of  the  given  numerators,  and  whose  denominator  is  the  least  com- 
mon multiple  of  the  given  denominators. 

1.  What  is  the  greatest  common  divisor  of  |,  T52,  and  ||? 

ANALYSIS.  The  greatest  common  divisor  of  5,  5,  and  15,  the  given 
numerators,  is  5.  The  least  common  multiple  of  6,  12,  and  16,  the 
given  denominators,  is  48.  Therefore  the  greatest  common  divisor 
of  the  given  fractions  is  ^\,  Ans.  (HI). 

PROOF. 

i  -*-.&-»•) 

T5TT  -T-  45g  =  4  >-  Prime  to  each  other. 

«+•*-•) 

198.  From  these  principles  and  illustrations,  we  derive  the 
following 


112  FRACTIONS. 

RULE.  Find  the  greatest  common  divisor  of  the  given  nume- 
rators for  a  new  numerator,  and  the  least  common  multiple  of  the 
given  denominators  for  a  new  denominator.  This  fraction  will  be 
the  greatest  common  divisor  sought. 

NOTE. — Whole  and  mixed  numbers  must  first  be  reduced  to  improper  fractions, 
and  all  fractions  to  their  lowest  terms. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  greatest  common  divisor  of  ^,  ^J,  and  ^f  ? 

Ans.  TJ7. 

2.  What  is  the  greatest  common  divisor  of  31,  1|,  and  |J? 

3.  What  is  the  greatest  common  divisor  of  4,  2|,  2|,  and  -g4^  ? 

Ans.  •£%. 

4.  What  is  the  greatest  common  divisor  of  109^  and  122|  ? 

5.  What  is   the   length  of  the   longest  measure   that  can  be 
exactly  contained  in  each  of  the  two  distances,  18|  feet  and  57^ 
feet  ?  Ans.  2-^  feet. 

6.  A  merchant  has  three  kinds  of  wine,  of  the  first  134f  gal- 
lons, of  the  second  128|  gallons,  of  the  third  115^  gallons;  he 
wishes  to   ship   the   same   in  full   casks  of  equal  size;  what  is 
the  least  number  he  can  use  without  mixing  the  different  kinds 
of  wine  ?     How  many  kegs  will  be  required  ?  Ans.  59. 

LEAST  COMMON  MULTIPLE  OF  FRACTIONS. 

199.  The  Least  Common  Multiple  of  two  or  more  fractions  is 
the  least  number  which  can  be  exactly  divided  by  each  of  them, 
giving  a  whole  number  for  a  quotient. 

200.  Since  in  performing  operations  in  division  of  fractions 
the  divisor  is  inverted,  it  is  evident  that  one  fraction  will  exactly 
contain  another  when  the  numerator  of  the  dividend  exactly  con- 
tains the  numerator  of  the  divisor,  and  the  denominator  oi'  the 
dividend  is  exactly  contained  in  the  denominator  of  the  divisor 
Hence, 

I.  A  fraction  is  a  multiple  of  a  given  fraction  when  its  nume- 
rator is  a  multiple  of  the  given  numerator,  and  its  denominator  is 
a  divisor  of  the  given  denominator.  And 


LEAST  COMMON  MULTIPLE.  113 

II.  A  fraction  is  a  common  multiple  of  two  or  more  given  frac- 
tions when  its  numerator  is  a  common  multiple  of  the  given  nume- 
rators, and   its   denominator  is  a  common  divisor  of  the  given 
denominators.     Therefore, 

III.  The  least  common  multiple  of  two  or  more  given  fractions 
is  a  fraction  whose  numerator  is  the  least  common  multiple  of  the 
given  numerators,  and  whose  denominator  is  the  greatest  common 
divisor  of  the  given  denominators. 

NOTE. — The  least  whole  number  that  will  exactly  contain  two  or  more  given 
fractions  in  their  lowest  terms,  is  the  least  common  multiple  of  their  numera- 
tors, (193,  Note  2). 

1.  What  is  the  least  common  multiple  of  |,  T52,  and  j|? 

ANALYSIS.  The  least  common  multiple  of  3,  5,  and  15,  the  given 
numerators,  is  15 ;  the  greatest  common  divisor  of  4,  12,  and  16,  the 
given  denominators,  is  4.  Hence,  the  least  common  multiple  of  the 
given  fractions  is  l(5  =  3£,  Ans.  (III). 

2O1.  From  these  principles  and  illustrations  we  derive  the 
following 

RULE.  Find  the  least  common  multiple  of  the  given  numerators 
for  a  new  numerator,  and  the  greatest  common  divisor  of  the  given 
denominators  for  a  new  denominator.  This  fraction  will  be  the 
least  common  multiple  sought. 

NOTE. — Mixed  numbers  and  integers  should  be  reduced  to  improper  fractions, 
and  all  fractions  to  their  lowest  terms,  before  applying  the  rule. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  least  common  multiple  of  f ,  T7^,  j|,  and  -f-g . 

Ans.  11 J. 

2.  What  is  the  least  common  multiple  of  -£%,  ||,  and  |$  ? 

3.  What  is  the  least  common  multiple  of  2f  J,  1|X,  and  -f^  ? 

4.  What  is  the  least  common  multiple  of  ^,  f ,  |,  |,  |,  |,  J,  f , 
and  -fy  ?  Ans.  2520. 

5.  The  driving  wheels  of  a  locomotive  are  15T5ff  feet  in  circum- 
ference, and  the  trucks  9|  feet  in  circumference.     What  distance 
must  the  train  move,  in  order  to  bring  the  wheel  and  truck  in  the 
same  relative  positions  as  at  starting  ?  Ans.  459|  feet. 

10*  H 


FRACTIONS. 


PROMISCUOUS    EXAMPLES. 

1.  Change  J  of  f  to  an  equivalent  fraction  having  135  for  its 
denominator.  Ans.  -ff^. 

2.  Reduce  |,  J,  |,  and  ||  to  equivalent  fractions,  whose  denom- 
inators shall  be  48. 

3.  Find  the  least  common  denominator  of  1^,  f  ,  2,  T7^,  |  of  i, 
I  off 

of  3  %  of  5 

5 


4.  The  sum  of  -  —  p-  ^  and  25        gt  is  equal  to  how  many  times 
their  difference  ?  Ans.  2. 

5.  The  less  of  two  numbers  is  j  —  7^,  and  their  difference  -~  ; 

s  ot  °s  T5 

what  is  the  greater  number  ?  -4»s.  34T4533 

6.  What  number  multiplied  by  f  of  f  X  3f  ,  will  produce  f  j  ? 


7.  Find  the  value  of         -»   X  +  (2  +  i)  -f-  (3  +  },) 

+  ^f--  ^«.  VA- 

5 

8.  What  number  diminished  by  the  difference  between  f  and  ^ 

of  itself,  leaves  a  remainder  of  144  ?  Ans.  283  J. 

9.  A  person  spending  -|,  |,  and  ^  of  his  money,  had  $119  left; 
how  much  had  he  at  first  ? 

10.  What  will  i  of  10|  cords  of  wood  cost,  at  2\  of  $42  per 
cord?  ~Ans.   $31^. 

11.  There  are  two  numbers  whose  difference  is  25T7-,  and  one 
number  is  f  of  the  other ;  what  are  the  numbers  ? 

Ans.  63|  and  89T\. 

12.  Divide  $2000  between  two  persons,  so  that  one  shall  have 
^  as  much  as  the  other.  Ans.  $1125  and  $875. 

13.  If  a  man  travel  4  miles  in  |  of  an  hour,  how  far  would  he 
travel  in  1^  hours  at  the  same  rate  ?  Ans.  10  miles. 

14.  At  $J  a  yard,  how  many  yards  of  silk  can  be  bought  for 
$10|? 

15.  How  many  bushels  of  oats  worth  $|  a  bushel,  will  pay  for 
J  of  a  barrel  of  flour  at  $7^  a  barrel  ? 


PROMISCUOUS  EXAMPLES. 

16.  If  I  of  a  bushel  of  barley  be  worth  |  of  a  bushel  of  corn, 
and  corn  be  worth  $f  per  bushel,  how  many  bushels  of  barley  will 
815  buy?  Ans.  18. 

17.  If  48  is  |  of  some  number,  what  is  |  of  the  same  number? 

18.  If  cloth  1|  yards  in  breadth  require  20^  yards  in  length  to 
make  a  certain   number  of  garments,  how  many  yards  in  length 
will  cloth  |  of  a  yard  wide  require  to  make  the  same  ? 

19.  A  gentleman  owning  |  of  an  iron  foundery,  sold  -i  of  his 
share  for  $2570| ;  how  much  was  the  whole  foundery  worth  ? 

Am.  $5141f 

20.  Suppose  the  cargo  of  a  vessel  to  be  worth  $10,000,  and  | 
of  f  of  y9^  of  the  vessel  be  worth  {  of  4-  of  1%  of  the  cargo ;  what 
is  the  whole  value  of  the  ship  and  cargo  ?  Ans.  $22000. 

21.  A  gentleman  divided  his  estate  among  his  three  sons  as  fol- 
lows :  to  the  first  he  gave  |  of  it ;  to  the  second  |  of  the  remain- 
der.    The  difference  between  the  portions  of  first  and  second  was 
$500.     What  was  the  whole  estate,  and  how  much  was  the  third 
son's  share  ?  f  Whole  estate,  $12000. 

^i/15.      *s 

{  Third  son's  share,  $2500. 

22.  If  7^  tons  of  hay  cost  $60,  how  many  tons  can  be  bought 
for  $78,  at  the  same  rate  ? 

23.  If  a  person  agree  to  do  a  job  of  work  in  30  days,  what  part 
of  it  ought  he  to  do  in  16^  days  ?  Ans.  ^. 

24.  A  father  divided  a  piece  of  land  among  his  three  sons ;  to 
the  first  he  gave  12^  acres,  to  the  second  |  of  the  whole,  and  to 
the  third  as  much  as  to  the  other  two ;  how  many  acres  did  the 
third  have?  Ans.  49  acres. 

25.  Iff  of  6  bushels  of  wheat  cost  $4^,  how  much  will  f  of  1 
bushel  cost? 

26.  A  man  engaging  in  trade  lost  f  of  his  money  invested,  after 
which  he  gained  $740,  when  he  had  $3500 ;  how  much  did  he 
lose?  Ans.  $1840. 

27.  A  cistern  being  full  of  water  sprung  a  leak,  and  before  it 
could  be  stopped,  |  of  the  water  ran  out,  but  |  as  much  ran  in  at 

the  same  time ;  what  part  of  the  cistern  was  emptied  ? 

Ans. 


116  FRACTIONS. 

28.  A  can  do  a  certain  piece  of  work  in  8  days,  and  B  oan  do 
the  same  in  6  days ;  in  what  time  can  both  together  do  it  ? 

Ans.  3 1  days. 

29.  A  merchant  sold  5  barrels  of  flour  for  $32  £,  which  was  § 
as  much  as  he   received   for  all  he  had  left,  at  $4  a  barrel ;  how 
many  barrels  in  all  did  he  sell  ?  Ana.  18. 

30.  What  is  the  least  number  of  gallons  of  wine,  expressed  by 
a  whole  number,  that  will  exactly  fill,  without  waste,  bottles  con- 
taining either  |,  J,  |,  or  f  gallons  ?  Ans.   60. 

31.  A,  B,  and  C  start  at  the  same  point  in  the  circumference 
of  a  circular  island,  and  travel  round  it  in  the  same  direction.     A 
makes  |  of  a  revolution  in  a  day,  B  T1y,  and  C  -58T.     In  how  many 
days  will  they  all  be  together  at  the  point  of  starting  ? 

Ans.  178 1  days. 

32.  Two  men  are  64 f  miles  apart,  and  travel  toward  each  other; 
when  they  meet  one  has  traveled  5£  miles  more  than  the  other; 
how  far  has  each  traveled  ? 

Ans.  One  29|  miles,  the  other  35&  miles. 

33.  There  are  two  numbers  whose  sum  is  ly1^,  and  whose  dif- 
ference is  |;  what  are  the  numbers?  Ans.  |  and  37^. 

34.  A,  B,  and  C  own  a  ferry  boat;  A  owns  j1^  of  the  boat, 
and  B  owns  775  of  the  boat  more  than  C.     What  shares  do  B  and 
C  own  respectively?  Ans.  B,  79^  ;  C,  J!J. 

35.  A  schoolboy  being  asked  how  many  dollars  he  had5  replied, 
that  if  his  money  be  multiplied  by  j|,  and  £  of  a  dollar  be  added 
to  the  product,  and  f  of  a  dollar  taken  from  the  sum,  this  remainder 
divided  by  •£§  would  be  equal  to  the  reciprocal  of  |  of  a  dollar. 
How  much  money  had  he  ? 

36.  If  a  certain  number  be  increased  by  If,  this  sum  diminished 
by  |,  this  remainder  multiplied  by  5|,  and  this  product  divided  by 
1*,  the  quotient  will  be  7£  ;  what  is  the  number?          Ans.  ||. 

37.  If  J  of  f  of  3£  times  l,be  multiplied  by  |,  the  product  di- 
vided by  §,  the  quotient  increased  by  4£,  and  the  sum  diminished  by 
£  of  iteelf,  what  will  the  remainder  be?  Ans. 


NOTATION  AND  NUMERATION. 

- 

DECIMAL  FRACTIONS. 
NOTATION  AND  NUMERATION. 

202.  A  Decimal  Fraction  is  one  or  more  of  the  decimal 
divisions  of  a  unit. 

NOTES. — 1.  The  word  decimal  is  derived  from  the  Latin  decem,  which  signi- 
fies ten. 

2.  Decimal  fractions  are  commonly  called  decimals. 

203.  In  the  formation  of  decimals,  a  simple  unit  is  divided 
into  ten  equal  parts,  forming  decimal  units  of  the  first  order,  or 
tenths,  each  tenth  is  divided  into  ten  equal  parts,  forming  decimal 
units  of  the  second  order,  or  hundredths;  and  so  on;  according  to 
the  following 

TABLE  OF  DECIMAL  UNITS. 

1  single  unit  equals  10  tenths ; 

1  tenth  "  10  hundredths; 

1  hundredth         "  10  thousandths ; 

1  thousandth        "  10  ten  thousandths 
etc.  etc. 

204.  In  the  notation  of  decimals  it  is  not  necessary  to  employ 
denominators  as   in   common   fractions;    for,  since  the   different 
orders  of  units  are  formed  upon  the  decimal  scale,  the  same  law 
of  local  value  as  governs  the  notation  of  simple  integral  numbers, 
(«57)>  enables  us  to  indicate  the  relations  of  decimals  by  place 
or  position. 

3O5.>  The  Decimal  Sign  (.)  is  always  placed  before  decimal 
figures  to  distinguish  them  from  integers.  It  is  commonly  called 
the  decimal  point.  When  placed  between  integers  and  decimals 
in  the  same  number,  is  sometimes  called  the  separatrix. 

2O6.  The  law  of  local  value,  extended  to  decimal  units,  as- 
signs the  first  place  at  the  right  of  the  decimal  sign  to  tenths ; 
the  second,  to  hundredths;  the  third,  to  thousandths;  and  so  on, 
as  shown  in  the  following 


118  DECIMALS. 


DECIMAL    NUMERATION    TABLE. 


ii         A 

§    i  £   S 

.11-        •&  jf^l! 

,    £  3  •  CQ       rS^^^fM 

i>vii    liilitl 

s-Ssllsl 111  Ilia 

043,3  S  °  ^  0^43  S  ^5  3 

c_ic^jMn-,h— ,^CL_iME_jr~LHri^ 


SMHHWHP^HKHH  WS 
5732754.  573256 


Integers.  Decimals. 

2OT.  The  denominator  of  a  decimal  fraction,  when  expressed, 
is  necessarily  10,  100,  1000,  or  some  power  of  10.  By  examining 
the  table  it  will  be  seen,  that  the  number  of  places  in  a  decimal  is 
equal  to  the  number  of  ciphers  required  to  express  its  denomi- 
nator. Thus,  tenths  occupy  the  first  place  at  the  right  of  units, 
and  the  denominator  of  j1^  has  one  cipher;  hundredths  in  the 
table  extend  two  places  from  units,  and  the  denominator  of  T^ 
has  two  ciphers;  and  so  on. 

2O8.  A  decimal  is  usually  read  as  expressing  a  certain  number 
of  decimal  units  of  the  lowest  order  contained  in  the  decimal. 
Thus,  5  tenths  and  4  hundredths,  or  .54,  may  be  read,  fifty-four 
hundredths.  For,  y5^  -f-  T^  =  T5<j40. 

2OO.  From  the  foregoing  explanations  and  illustrations  we 
derive  the  following 

PRINCIPLES    OF   DECIMAL   NOTATION    AND    NUMERATION. 

I.  Decimals  are  governed  by  the  same  law  of  local  value  that 
governs  the  notation  of  integers. 

II.  The  different  orders  of  decimal  units  decrease  from  left  to 
right,  and  increase  from  right  to  left,  in  a  tenfold  ratio. 


NOTATION  AND  NUMERATION.  H9 

III.  The  value  of  any  decimal  figure  depends  upon  the  place 
it  occupies  at  the  right  of  the  decimal  sign. 

IV.  Prefixing  a  cipher  to  a  decimal  diminishes  its  value  ten- 
fold, since  it  removes  every  decimal  figure  one  place  to  the  right. 

V.  Annexing  a  cipher  to  a  decimal  does  not  alter  its  value, 
since  it  does  not  change  the  place  of  any  figure  in  the  decimal. 

VI.  The  denominator  of  a  decimal,  when    expressed,  is  the 
unit,  1,  with  as  many  ciphers  annexed  as  there  are  places  in  the 
decimal. 

VII.  To  read  a  decimal  requires  two  numerations;  first,  from 
units,  to  find  the  name  of  the  denominator ;  second,  towards  units, 
to  find  the  value  of  tho  numerator. 

21O.  Having  analyzed  all  the  principles  upon  which  the 
writing  and  reading  of  decimals  depend,  we  will  now  present  these 
principles  in  the  form  of  rules. 

RULE    FOR   DECIMAL   NOTATION. 

* 

I.  Write   the  decimal   the   same  as  a  whole   number,  placing 
ciphers  in  the  place  of  vacant  orders,  to  give  each  significant  figure 
its  true  local  value. 

II.  Place  the  decimal  point  before  the  first  figure. 

RULE   FOR   DECIMAL    NUMERATION. 

I.  Numerate  from  the  decimal  point,  to  (Jeter mine  the  denomi- 
nator. 

II.  Numerate  towards  the  decimal  point,  to  determine  the  nu- 
merator. 

III.  Read  the  decimal  as  a  whole  number,  giving  it  the  name 
of  its  lowest  decimal  unit,  or  right  hand  figure. 

EXAMPLES    FOR   PRACTICE. 

Express  the  following  decimals  by  figures,  according  to  the 
decimal  notation. 

1.  Five  tenths.  Ans.  .5. 

2.  Thirty-six  hundredths.  Ans.  .36. 

3.  Seventy -five  ten-thousandths.  Ana.  .0075. 


120 


DECIMALS. 


4.  Four  hundred  ninety-six  thousandths. 

5.  Three  hundred  twenty-five  ten-thousandths. 
6    One  millionth. 

7.  Seventy -four  ten-million  ths. 

8.  Four  hundred  thirty-seven  thousand  five  hundred  forty- 
nine  millionths. 

9.  Three  million  forty  thousand  ten  ten-millionths. 

10.  Twenty-four  hundred-millionths. 

11.  Eight  thousand  six  hundred  forty-five  hundred-thousandths. 

12.  Four  hundred  ninety-five  million  seven  hundred  five  thou- 
sand forty-eight  billionths. 

13.  Ninety-nine  thousand  nine  ten-billionths. 

14.  Four  million  seven  hundred  thirty-five  thousand  nine  hun- 
dred one  hundred-millionths. 

15.  One  trillionth. 

16.  One  trillion  one  billion  one  million  one  thousand  one  ten- 
trillionths. 

17.  Eight  hundred  forty-one  million  five  hundred  sixty-three 
thousand  four  hundred  thirty-six  trillionths. 

18.  Nine  quintillionths. 

Express  the  following  fractions  and  mixed  numbers  decimally : 


19.  TV  Ans.  .3. 

20. 


22. 


24.  T1 

Head  the  following  numbers 

31.  .24. 

32.  .075. 

33.  .503. 

34.  .00725. 

35.  .40000004. 

36.  .0000256. 

37.  .0010075. 


25.  46TV 
26. 

27. 
28. 
29. 
30. 


38. 
39. 
40. 
41. 
42. 
43. 
44. 


Ans.  46.4. 


8.25. 
75.368. 
42.0637. 

8.0074. 
30.4075. 
26.00005. 
100.00000001. 


REDUCTION,  121 

REDUCTION. 
CASE  I. 

211.  To  reduce  decimals  to  a  common  denominator. 
1.  Reduce  .5,  .24,  .7836  and  .375  to  a  common  denominator. 

ANALYSIS.     A  common  denominator  must  contain 

OFERA11ON. 

'SOOO  as  manv  decimal  places  as  are  equal  to  the  greatest 
2400  number  of  decimal  figures  in  any  of  the  given  deci- 
7g3(J  mals.  We  find  that  the  third  number  contains  four 

.3750          decimal  places,  and  hence  10000  must  be  a  common 
denominator.     As  annexing  ciphers  to  decimals  does 

not  alter  their  value,  we  give  to  each  number  four  decimal  places,  by 

annexing  ciphers,  and  thus  reduce  the  given  decimals  to  a  common 

denominator.     Hence, 

RULE.      Give   to  each  number  the  same  number  of  decimal 

places,  by  annexing  ciphers. 

NOTES. — 1.  If  the  numbers  be  reduced  to  the  denominator  of  that  one  of  the 
given  numbers  having  the  greatest  number  of  decimal  places,  they  will  have 
their  least  common  decimal  denominator. 

2.  An  integer  may  readily  be  reduced  to  decimals  by  placing  the  decimal 
point  after  units,  and  annexing  ciphers  ;  one  cipher  reducing  it  to  tenths,  two 
ciphers  to  hundredths,  three  ciphers  to  thousandths,  and  so  on. 

EXAMPLES   FOR   PRACTICE. 

1.  Reduce  .18,  .456,  .0075,  .000001,  .05,  .3789,  .5943786,  and 
.001  to  their  least  common  denominator. 

2.  Reduce  12  thousandths,  185  million ths,  936  billion  ths,  and 
7  trillionths  to  their  least  common  denominator. 

3.  Reduce  57.3,  900,  4.7555,  and  100.000001  to  their  least 
common  denominator. 

CASE   II. 

212.  To  reduce  a  decimal  to  a  common  fraction. 
1.  Reduce  .375  to  an  equivalent  common  fraction. 

OPERATION.  ANALYSIS.    Writing  the  decimal 

Q-r  _     375          3  figures,  .375,  over  the  common  de- 

55  TS  <J<?  =  s-  nominator,  1000,  we  have  T3j%  =  f . 
Hence, 

11 


122  DECIMALS.  ; 

RULE.      Omit  the  decimal  point,  and  supply  the  proper  denomi- 
nator. 

EXAMPLES    FOR    PRACTICE. 

1.  Reduce  .75  to  a  common  fraction.  Ans.  |. 

2.  Reduce  .625  to  a  common  fraction.  Ans.  |. 

3.  Reduce  .12  to  a  common  fraction. 

4.  Reduce  .68  to  a  common  fraction. 
'5.  Reduce  .5625  to  a  common  fraction. 

6.  Reduce  .024  to  a  common  fraction.  Ans.  Tj^. 

7.  Reduce  .00032  to  a  common  fraction.  Ans. 

8.  Reduce  .002624  to  a  common  fraction.  Ans. 

9.  Reduce  ,13|  to  a  common  fraction. 

OPERATION. 


NOTE.  —  The  decimal  .13J  may  properly  be  called  a  complex  decimal. 

10.  Reduce  .57|  to  a  common  fraction.  Ans.  %, 

11.  Reduce  .66|  to  a  common  fraction.  Ans.  f  . 

12.  Reduce  .444|  to  a  common  fraction. 

13.  Reduce  .024|  to  a  common  fraction.  Ans.  Tf  J$. 

14.  Reduce  ,984|  to  a  common  fraction. 

15.  Express  7.4  by  an  integer  and  a  common  fraction. 

Ans.  7|. 

16.  Express  24.74  by  an  integer  and  a  common  fraction. 

17.  Reduce  2.1875  to  an  improper  fraction.  Ans.  ||. 

18.  Reduce  1.64  to  an  improper  fraction. 

19.  Reduce  7.496  to  an  improper  fraction.  Ans.  |||. 

CASE   III. 

313.  To  reduce  a  common  fraction  to  a  decimal. 
1.  Reduce  |  to  its  equivalent  decimal. 

FIRST    OPERATION.  ANALYSIS.       We     first 

the  8ame  number  of  ciphers  to 


_G25    . 

i  UT)U  both  tcrmg  of  the  fraction  ;  this 

does  not  alter  its  value,  (174, 


DEDUCTION.  123 

SECOND  OPERATION.  Ill) ;  we  then  divide  both  re- 

8  ")  5  000  suiting  terms  by  8,  the  signifi- 

cant figure  of  the  denominator, 
'"  and  obtain  the  decimal  denom- 

inator, 1000.     Omitting  the  denominator,  and  prefixing  the  sign,  we 
have  the  equivalent  decimal,  .625. 

In  the  second  operation,  we  omit  the  intermediate  steps,  and  obtain 
the  result,  practically,  by  annexing  the  three  ciphers  to  the  purne- 
rator,  5,  and  dividing  the  result  by  the  denominator,  8. 

2.  Reduce  T|3  to  a  decimal. 

OPERATION.  ANALYSIS.     Dividing  as  in  the  former  ex- 

125  )  3.000          ample,  we  obtain  a  quotient  of  2  figures,  24.. 

Q24          But  since  3  ciphers  have  been  annexed  to  the 

numerator,  3,  there  must  be  three  places  in  the 

required  decimal ;  hence  we  prefix  1  cipher  to  the  quotient  figures, 
24.     The  reason  of  this  is  shown  also  in  the  following  operation. 

3.  3000  24         _    09-1 

J^-5  —  725UUU  —  TSUU  —  'U^ 

214U    From  these  illustrations  we  derive  the  following 
RULE.     I.  Annex  ciphers  to  the  numerator,  and  divide  ~by  the 

denominator. 

II.  Point  off  as  many  decimal  places  in  the  result  as  arc  equal 

to  the  number  of  ciphers  annexed. 

NOTE.-  If  the  division  is  not  exact  when  a  sufficient  number  of  decimal 
figures  have  been  obtained,  the  sign,  +,  may  bo  annexed  to  the  decimal  to  indi- 
cate that  there  is  still  a  remainder.  When  this  remainder  is  such  that  the  next 
decimal  figure  would  be  5  or  greater  than  5,  the  last  figure  of  the  terminated 
decimal  may  be  increased  by  1,  and  the  sign,  — ,  annexed.  And  in  general,  + 
denotes  that  the  written  decimal  is  too  small,  and  —  denotes  that  the  written 
decimal  is  too  large ;  the  error  always  being  less  than  one  half  of  a  unit  in  the 
last  place  of  the  decimal. 

EXAMPLES   FOR   PRACTICE. 

1.  Reduce  |  to  a  decimal.  Ans.  .75. 

2.  Reduce  T5ff  to  a  decimal.  Ans.  .3125. 

3.  Reduce  J  to  a  decimal. 

4.  Reduce  ^|  to  a  decimal. 

5.  Reduce  -if  to  a  decimal.  % 

6.  Reduce  ^  to  a  decimal.  Ans.   .04. 

7.  Reduce  V/s  to  a  decimal.  Ans    .068. 

• 


124  DECIMALS. 

8.  Reduce  ^f  to  a  decimal.  Ans.  .59375 

9.  Reduce  y^g3^  to  a  decimal. 

10.  Reduce  37?  to  a  decimal.  Ans.   .29167 — . 

11.  Reduce  j9^  to  a  decimal. 

12.  Reduce  jf  to  a  decimal.  Ans.  .767857  +  . 

13.  Reduce  7^  to  the  decimal  form.  Ans.  7.125. 

14.  Reduce  56^  to  the  decimal  form.          Ans.  56.078125. 

15.  Reduce  32|  to  the  decimal  form. 

16.  Reduce  .24^  to  a  simple  decimal. 

17.  Reduce  5.78|-Q  to  a  simple  decimal. 

18.  Reduce  .3T^  to  a  simple  decimal.  Ans.  .30088. 

19.  Reduce  4.032-  to  a  simple  decimal.  Ans.  4.008. 

20.  Reduce  .30T||Ji^  to  a  simple  decimal. 


ADDITION. 

21*1.  Since  the  same  law  of  local  value  extends  both  to  the 
right  and  left  of  units'  place;  that  is,  since  decimals  and  simple 
integers  increase  and  decrease  uniformly  by  the  scale  of  ten,  it  is 
evident  that  decimals  may  be  added,  subtracted,  multiplied  and 
divided  in  the  same  manner  as  integers. 

216.     1.  What  is  the  sum  of  4.75,  .246,  37.56  and  12.248  ? 

OPERATION.          ANALYSIS.     We  write  the  numbers  so  that  units  of 
4.75  like  orders,  whether  integral  or  decimal,  shall  stand 

.246  in  the  same  columns ;  that  is,  units  under  units,  tenths 

37.56  under  tenths,  etc.     This  brings  the  decimal  points 

12.248  directly  under  each  other.     Commencing  at  the  right 

54  804  hand,  we  add  each  column  separately,  carrying  1  for 

every  ten,  according  to  the  decimal  scale  ;  and  in  the 
result  we  place  the  decimal  point  between  units  and  tenths,  or  directly 
under  the  decimal  points  in  the  numbers  added.  Hence  the  fol- 
lowing 

RULE.  I.  Write  the  numbers  so  that  the  decimal  points  shall 
stand  directly  under  each  other. 

II.  Add  as  in  whole  numbers,  and  place  the  decimal  point,  in 
ihe  result,  directly  under  the  points  in  the  numbers  added. 


ADDITION.  125 

EXAMPLES    FOR   PRACTICE. 

1.  Add  .375,  .24,  .536,  .78567,  .4637,  and  .57439. 

Ans.  2.97476. 

2.  Add  5.3756,  85.473,  9.2,  46.37859,  and  45.248377. 

An*.  191.675567. 

3.  Add  .5,  .37,  .489,  .6372,  .47856,  and  .02524. 

4.  Add  .46},  .325J,  .16^,  and  .275TV       Ans.  1.2296625. 

5.  Add  4.6£,  7.32 3^,  5.3784^,  and  2.64878 j. 

6.  Add  4.3785,  2|,  5f,  and  12.4872.  Ans.  24.9609-f . 

7.  What  is  the  sum  of  137  thousandths,  435  thousandths,  836 
thousandths,  937  thousandths,  and  496  thousandths  ? 

Ans.  2.841. 

8.  What  is  the  sum  of  one  hundred  two  ten- thousandths,  thir- 
teen thousand  four  hundred  twenty-six  hundred  thousandths,  five 
hundred  sixty-seven  millionths,  three  millionths,  and  twenty-four 
thousand  seven  hundred-thousandths  ? 

9.  A  farm  has  five  corners;  from  the  first  to  the  second  is  34.72 
rods;  from  the  second  to  the  third,  48.44  rods;  from  the  third  to 
the  fourth,  152.17  rods;  from  the  fourth  to  the  fifth,  95.36  rods; 
and  from  the  fifth  to  the  first,  56.18  rods.     What  is  the  whole 
distance  around  the  farm  ? 

10.  Find  the  sum  of  ||,  ^5ff,  T3^,  and  7773  in  decimals,  correct 
to  the  fourth  place.  Ans.  .6666 -f-- 

NOTE. — In  the  reduction  of  each  fraction,  carry  the  decimal  to  at  least  the 
fifth  place,  in  order  to  insure  accuracy  in  the  fourth  place. 

11.  A  man  owns  4  city  lots,  containing  16Te^  rods,  15T27  rods, 
181?-  rods,  and  14?75  rods  of  land,  respectively;  how  many  rods 
in  all? 

12.  What  is  the  sum  of  4  4  decimal  units  of  the  first  order,  2| 
of  the  second  order,  9^  of  the  third  order,  and  3,TV  of  the  fourth 
order?  Ans.  .486929. 

13.  What  is  the  approximate  sum  of  1  decimal  unit  of  the  first 
order,  4  of  a  unit  of  the  second  order,  1  of  a  unit  of  the   third 
order,  ^  of  a  unit  of  the  fourth  order,  1  of  a  unit  of  the  fifth  order, 
J  of  a  unit  of  the  sixth  order,  and  %  of  a  unit  of  the  seventh  order  ? 

Ans.  .1053605143—. 
11* 


126  DECIMALS. 

i 

SUBTRACTION. 
1.  From  4.156  take  .5783. 

ANALYSIS.  We  write  the  given  numbers  as  in  addi- 
tion, reduce  the  decimals  to  a  common  denominator, 
and  subtract  as  in  integers.  Or,  we  may,  in  practice, 
omit  the  ciphers  necessary  to  reduce  the  decimals  to  a 
common  denominator,  and  merely  conceive  them  to  be 
annexed,  subtracting  as  otherwise.  Hence  the  fol- 
lowing 

3^5777 

RULE.  I.  Write  the  numbers  so  that  the  decimal' points  shaft 
vtand  directly  under  each  other. 

II.  Subtract  as  in  whole  numbers,  and  place  the  decimal  point 
in  the  result  directly  under  the  points  in  the  given  numbers. 

EXAMPLES    FOR    PRACTICE. 

00  (20  (30 

Minuend,       .9876  48.3676  36.5 

subtrahend,    .3598  23.98  35.875632 

Remainder,      .6278  24.3876  .624368 

4.  From  37.456  take  24.367.  Ans.  13.089. 

5.  From  1.0066  take  .15. 

6.  From  1000  take  .001.  Ans.  999.999. 

7.  From  36|  take  22£f .  Ans.  14.27. 

8.  From  .56J  take  .55{f  f . 

9.  From  7|  take  579ff.  Ans.   1.7708  +  . 

10.  From  114  take  ?ja. 

11.  From  one  take  one  trillionth.          Ans.  .999999999999. 

12.  A  speculator  having  57436  acres  of  land,  sold  at  different 
times  536.74  acres,  1756.19  acres,  3678.47  acres,  9572.15  acres, 
7536.59    acres,    and   4785.94    acres;    how    much    land    has   he 
remaining? 

13.  Find  the  difference  between  f|||i  and  iff  ^f,  correct  to 
the  Efth  decimal  place.  Ans.  4.17298-f . 


MULTIPLICATION.  127 


MULTIPLICATION. 

In  multiplication  of  decimals,  the  location  of  the  decimal 
point  in  the  product  depends  upon  the  following  principles : 

I.  The  number  of  ciphers  in  the  denominator  of  a  decimal  is 
equal  to  the  number  of  decimal  places,  (2OO,  "VI). 

II.  If  two  decimals,  in  the  fractional  form,  be  multiplied  to- 
gether, the  denominator  of  the  product  must   contain  as  many 
ciphers  as  there  are  decimal  places  in  both  factors.     Therefore, 

III.  The  product  of  two  decimals,  expressed  in  the  decimal 
form,  must  contain  as  many  decimal  places  as  there  are  decimals 
in  both  factors. 

1.  Multiply  .45  by  .7. 

OPERATION.  ANALYSIS.   We  first  multiply 

45  as   in  whole   numbers ;   then, 

since  the  multiplicand  has  2 
„..  p.  decimal  places  and  the  multi- 

plier 1,  we  point  off  2  +  1  =  3 
PROOF.  decimal  places  in  the  product, 

=  -315       (m)-   The  reason  of  this  is 

further  illustrated  in  the  proof, 
a  method  applicable  to  all  similar  cases. 

219.  Hence  the  following 

RULE.  Multiply  as  in  whole  numbers,  and  from  the  right  hand 
of  the  product  point  off  as  many  figures  for  decimals  as  there  are 
decimal  places  in  loth  factors. 

NOTES. —  1.  If  there  be  not  as  many  figures  in  the  product  as  there  are  deci- 
mals in  both  factors,  supply  the  deficiency  by  prefixing  ciphers. 

2.  To  multiply  a  decimal  by  10,  100,  1000,  etc.,  remove  the  point  as  many 
places  to  the  right  as  there  are  ciphers  on  the  right  of  the  multiplier. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  .75  by  .41.  Ans.  .3075. 

2.  Multiply  .436  by  .24. 

3.  Multiply  5.75  by  .35.  Ans.  2.0125. 

4.  Multiply  .756  by  .025.  'An*.  .0189. 

5.  Multiply  3. 784  by  2.475. 


128  DECIMALS. 

t 

6.  Multiply  7.23  by  .0156.  Ant.  .112788. 

7.  Multiply  .0075  by  .005.  Am.  .0000375. 

8.  Multiply  324  by  .324. 

9.  Multiply  75.64  by  .225. 

10.  Multiply  5.728  by  100.  Ans.  572.8. 

11.  Multiply  .36  by  1000. 

J2.  Multiply  .000001  by  1000000. 

13.  Multiply  .576  by  100000. 

14.  Multiply  7|  by  5£.  Ans.  42.625. 

15.  Multiply  .63$  by  24. 

16.  Multiply  4/g  by  7^.  Ans.  31.74. 

17.  Find  the  value  of  3.425  x  1.265  x  64.    Ans.  277.288. 

18.  Find  the  value  of  32  x  .57825  x  .25. 

19.  Find  the  value  of  18.375  x  5.7  x  1.001. 

Ans.  104.8422375. 

20.  If  a   cubic   foot   of  granite  weigh    168.48  pounds,  what 
will  be  the  weight  of  a  granite  block  that  contains  27|  cubic 
feet? 

21.  When  a  bushel  of  corn  is  worth  2.8  bushels  of  oats,  how 
many  bushels  of  oats  must  be  given  in  exchange  for  36  bushels 
of  corn  and  48  bushels  of  oats  ?  Ans.  148.8. 


CONTRACTED    MULTIPLICATION. 

22O.  To  obtain  a  given  number  of  decimal  places  in 
the  product. 

It  is  frequently  the  case  in  multiplication,  that  a  greater  number 
of  decimal  figures  is  obtained  in  the  product,  than  is  necessary 
for  practical  accuracy.  This  may  be  avoided  by  contracting  each 
partial  product  to  the  required  number  of  decimal  places. 

To  investigate  the  principles  of  this  method,  let  us  take  the  two 
decimals  .12345  and  .54321,  and  having  reversed  the  order  of  the 
digits  in  the  latter,  and  written  it  under  the  former,  multiply  each 
figure  of  the  direct  number  by  the  figure  below  in  the  reversed  num- 
ber, placing  the  products  with  like  orders  of  units  in  the  same  column, 
thus: 


CONTRACTED  MULTIPLICATION. 

.12345  direct     =  .12345 
.54321  reversed  =  12345. 


.000025  =  .00005  X  .5 
.000016  =  .0004    x  .04 
.000009  =  .003      X  .003 
.000004  =  .02        X  .0002 
.000001  =  .1          X  .00001. 

In  this  operation  we  perceive  that  all  the  products  are  of  the  same 
.order;  and  this  must  always  be,  whether  the  numbers  used  be  frac- 
tional, integral,  or  mixed.  For,  as  we  proceed  from  right  to  left  in 
the  multiplication,  we  pass  regularly  from  lower  to  higher  orders  in 
the  direct  number,  and  from  higher  to  lower  in  the  reversed  number. 
Hence 

221.  If  one  number  be  written  under  another  with  the  order 
of  its  digits  reversed,  and  each  figure  of  the  reversed  number  be 
multiplied  by  the  figure  above  it  in  the  direct  number,  the  prod- 
ucts will  all  be  of  the  same  order  of  units. 

1.  Multiply  4.78567  by  3.25765,  retaining  only  3  decimal 
places  in  the  product. 

OPERATION.  ANALYSIS.    Since  the  product 

4.  78~P7  °^  an^  ^§ure  kv  units  is  of  the 

W~£i9  3  same  order  as  the  figure  multi- 

plied, (82,  II,)  we  write  3,  the 
units  of  the  multiplier,  under 
9jjJ  —     478  X  2  -f  1          5    the  third  decimal  figure  of 

OQQ  __          4-7    V    0      I     A 

*  -  _T_  r  the  multiplicand,  and  the  lowest, 
n  v  r>  _i_  o  order  to  be  retained  in  the  pro- 
duct ;  and  the  other  figures  of 


15.589  ±,  Ans.  the  multiplier  we 


inverted  order,  extending  to  the  left.  Then,  since  the  product  of  3 
and  5  is  of  the  third  order,  or  thousandths,  the  products  of  the  other 
corresponding  figures  at  the  left,  2  and  8,  5  and  7,  7  and  4,  etc.,  will 
be  thousandths ;  and  we  therefore  multiply  each  figure  of  the 
multiplier  by  the  figures  above  and  to  the  left  of  it  in  the  multipli- 
cand, carrying  from  the  rejected  figures  of  the  multiplicand,  as  fol- 
lows :  3  times  6  are  18,  and  as  this  is  nearer  2  units  than  one  of  the 
next  higher  order,  we  must  carry  2  to  the  first  contracted  product ;  3 
times  5  are  15,  and  2  to  be  carried  are  17  ;  writing  the  7  under  the 
3,  and  multiplying  the  other  figures  at  the  left  in  the  usual  manner, 


130  DECIMALS.  ; 

we  obtain  14357  for  the  first  partial  product.  Then,  beginning  with 
the  next  figure  of  the.  multiplier,  2  times  5  are  10,  which  gives  1  to 
be  carried  to  the  second  partial  product ;  2  times  8  are  16,  and  1  to  be 
carried  are  17  ;  writing  the  7  under  the  first  figure  of  the  former  pro- 
duct, and  multiplying  the  remaining  left-hand  figures  of  the  mul- 
tiplicand, we  obtain  957  for  the  second  partial  product.  Then,  5 
times  8  are  40,  which  gives  4  to  be  carried  to  the  third  partial  pro- 
duct ;  5  times  7  are  35  and  4  are  39  ;  writing  the  9  in  the  first  column 
of  the  products,  and  proceeding  as  in  the  former  steps,  we  obtain  239 
for  the  third  partial  product.  Next,  multiplying  by  7  in  the  same 
manner,  we  obtain  33  for  the  fourth  partial  product.  Lastly,  begin- 
ning 2  places  to  the  right  in  the  multiplicand,  6  times  7  are  42 ;  6 
times  4  are  24,  and  4  are  28,  which  gives  3  to  be  carried  to  the  fifth 
partial  product;  6  times  0  is  0,  and  3  to  be  carried  are  3,  which  we 
write  for  the  last  partial  product.  Adding  the  several  partial  pro- 
ducts, and  pointing  off  3  decimal  places,  we  have  15.589,  the  required 
product. 

222.  From  these  principles  and  illustrations  we  derive  the 
following 

RULE.  I.  Write  the  multiplier  with  the  order  of  its  figures 
reversed,  and  with  the  units'  place  under  that  figure  of  the  multi- 
plicand which  is  the  lowest  decimal  to  be  retained  in  the  product. 

II.  Find  the  product  of  each  figure  of  the  multiplier  by  the 
figures  above  and  to  the  left  of  it  in  the  multiplicand,  increasing 
each  partial  product  by  as  many  units  as  would  have  been  carried 
from  the  rejected p art  of  the  multiplicand,  and  one  more  when  the 
highest  figure  in  the  rejected  part  of  any  product  is  5  or  greater 
than  5  ',  and  write  these  partial  products  with  the  lowest  figure  of 
each  in  the  same  column. 

III.  Add  the  partial  products,  and  from  the  right  hand  of  the 
result  point  off  the  required  number  of  decimal  figures. 

NOTES. — 1.  In  obtaining  the  number  to  be  carried  to  each  contracted  partial 
product,  it  is  generally  necessary  to  multiply  (mentally)  only  one  figure  at  the 
right  of  the  figure  above  the  multiplying  figure;  but  when  the  figures  are  large, 
the  multiplication  should  commence  at  lenst  two  places  to  the  right. 

2.  Observe,  that  when  the  number  of  units  in  the  highest  order  of  the  rejected 
part  of  the  product  is  between     5   and  15,  carry  1;  if  between  15  and  25  carry 
2 ;  if  between  25  and  35  carry  3 ;  and  so  on. 

3.  There  is  always  a  liability  to  an  error  of  one  or  two  units  in  the  last  place; 
and  as  the  answer  may  be  either  too  great  or  too  small  by  the  amount  of  thii 


CONTRACTED  MULTIPLICATION. 

error,  the  uncertainty  may  be  indicated  by  the   double  sign,  i,  read,  plus,  or 
minus,  and  placed  after  the  product. 

4.  When   the  number  of  decimal  places  in  the   multiplicand  is  less  than  the 
number  to  be  retained  in  the  product,  supply  the  deficiency  by  annexing  ciphers. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  236.45  by  32.46357,  retaining  2  decimal  places, 
and  2.563789  by  .0347263,  retaining  6  decimal  places  in  the 
product. 

OPERATION.  OPERATION. 

236.450  2.563789 

75364.23  362  7430. 


76914 

47290  10255 

9458  1795 

1419  51 

71  15 

12  1 


2  .089031 


7676.02  rb 

2.  Multiply  36.275  by  4.3678,  retaining  1  decimal  place  in  the 
product.  Ans.  158.4  db. 

3.  Multiply  .24367  by  36.75,  retaining  2  decimal  places  in  the 
product. 

4.  Multiply  4256.785  by  .00564,  rejecting  all  beyond  the  third 
decimal  place  in  the  product.  Ans.  24.008  ±. 

5.  Multiply  357.84327  by  1.007806,  retaining  4  decimal  places 
in  the  product. 

6.  Multiply  400.756  by  1.367583,  retaining  2  decimal  places  in 
the  product.  Ans.  548.07  ±. 

7.  Multiply  432.5672  by  1.0666666,  retaining  3  decimal  places 
in  the  product. 

8.  Multiply  48.4367  by  2^,  extending  the  product  to  three 
decimal  places.  Ans.  103.418  rfc. 

9.  Multiply  7-j-fj  by  3|||,  extending  the  product  to  three 
decimal  places. 

10.  The  first  satellite  of  Uranus  moves  in  its  orbit  142.8373  + 


132  DECIMALS. 

degrees  in  1  day;  find  how  many  degrees  it  will  move  in  2.52035 
days,  carrying  the  answer  to  two  decimal  places. 

Ans.  360.00  degrees. 

11.  A  gallon  of  distilled  water  weighs  8.33888  pounds ;  how 
many  pounds  in  35.8756  gallons  ?          Ans.  299.16  d=  pounds. 

12.  One  French  metre  is  equal  to  1.09356959  English  yards; 
how  many  yards  in  478.7862  metres.        Ans.  523.58  ±  yards. 

13.  The  polar  radius  of  the  earth  is  6356078.96  metres,  and  the 
equatorial  radius,   6377397.6   metres;  find  the  two  radii,  and  their 
difference,  to  the  nearest  hundredth  of  a  mile,  1  metre  being  equal 
to  0.000621346  of  a  mile. 


DIVISION. 

223.    In  division  of  decimals  the  location  of  the  decimal 
point  in  the  quotient  depends  upon  the  following  principles : 

I.  If  one  decimal  number  in  the  fractional  form  be  divided  by 
another  also  in  the  fractional  form,  the  denominator  of  the  quotient 
must  contain  as  many  ciphers  as  the  number  of  ciphers  in  the  de- 
nominator of  the  dividend  exceeds  the  number  in  the  denominator 
of  the  divisor.     Therefore, 

II.  The  quotient  of  one  number  divided  by  another  in  the  deci- 
mal form  must  contain  as  many  decimal  places  as  the  number  of 
decimal  places  in  the  dividend  exceed  the  number  in  the  divisor. 

1.  Divide  34.368  by  5.37. 

ANALYSIS.     We  first  divide  as 

OFiiKATION. 

c  0-  N  0^  or.0  /  n  A  in  whole  numbers;  then,  since  the 

O.o7   )  o^doo  (  D.4  ,.  .,      T  ,       01-      -i    ! 

g2  2?  dividend  has  3  decimal  places  and 

the  divisor  2,  we  point  off  3  —  2 
=  1  decimal  place  in  the  quotient, 
(II).  The  correctness  of  the  work 
is  shown  in  the  proof,  where  the 
dividend  and  divisor  are  written  as 

Jritv/Ul1  • 

common  fractions.     For,  when  we 

TflUC    x  557  =  H  =  6-4        have  canceled  the  denominator  of 

the  divisor  from  the  denominator 
of  the  dividend,  the  denominator  of  the  quotient  must  contain  aa 


DIVISION.  133 

many  ciphers  as  the  number  in  the  dividend  exceeds  those  in  the 
divisor. 

S834L    Hence  the  following 

RULE.  Divide  as  in  whole  numbers,  and  from  the  right  hand 
of  the  quotient  point  off  as  many  places  for  decimals  as  the  decimal 
places  in  the  dividend  exceed  those  in  the  divisor. 

NOTES. — 1.  If  the  number  of  figures  in  the  quotient  be  less  than  the  excess  of 
the  decimal  places  in  the  dividend  over  those  in  the  divisor,  the  deficiency  must 
be  supplied  by  prefixing  ciphers. 

2.  If  there  be  a  remainder  after  dividing  the  dividend,  annex  ciphers,  and 
continue  the  division  :  the  ciphers  annexed  are  decimals  of  the  dividend. 

3.  The  dividend  should  always  contain  at  least  as  many  decimal  places  as  the 
divisor,  before  commencing  the  division;  the  quotient  figures  will  then  be  inte- 
gers till  all  the  decimals  of  the  dividend  have  been  used  in  the  partial  dividends. 

4.  To  divide  a  decimal  by  10,  100, 1000,  etc.,  remove  the  point  as  many  places 
to  the  left  as  there  are  ciphers  on  the  right  of  the  divisor. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  9.6188  by  3.46.  Ans.  2.78. 

2.  Divide  46.1975  by  54.35.  Ant.  .85. 

3.  Divide  .014274  by  .061.  Ans.  .234. 

4.  Divide  .952  by  4.76. 

5.  Divide  345.15  by  .075.  Ans.  4602. 

6.  Divide  .8  by  476.3.  Ans.  .001679  +  . 

7.  Divide  .0026  by  .003. 

8.  Divide  3.6  by  .00006!  Ans.  60000. 

9.  Divide  3  by  450. 

10.  Divide  75  by  10000. 

11.  Divide  4.36  by  100000. 

12.  Divide  .1 'by  .12. 

13.  Divide  645.5  by  1000. 

14.  If  25  men  build  154.125  rods  offence  in  a  day,  how  much 
does  each  man  build  ? 

15.  How  many  coats  can  be  made  from   16.2  yards  of  cloth, 
allowing  2.7  yards  for  each  coat  ? 

16.  If  a  man  travel  36.34  miles  a  day,  how  long  will  it  take 
him  to  travel  674  miles  ?  Ans.  18.547-|-days. 

17.  How  many  revolutions  will  a  wheel  14.25  feet  in  circum- 
ference make  in  going  a  distance  of  1  mile  or  5280  feet  ? 


134 


DECIMALS. 


CONTRACTED    DIVISION. 

To  obtain  a  given  number  of  decimal  places  in 
the  quotient. 

In  division,  the  products  of  the  divisor  by  the  several  quotient 
figures  maybe  contracted,  as  in  multiplication,  by  rejecting  at  each 
step  the  unnecessary  figures  of  the  divisor,  (22O). 

1.  Divide  790.755197  by  32.4687,  extending  the  quotient  to 
two  decimal  places. 

FIRST    CONTRACTED    METHOD.  COMMON    METHOD. 


32.4687)790.755197(24.35 
6494 

1413 
1299 


32.4687  )  790.7  55198  (  24.35 
649  3  74 

141  3'811 

129  874* 


114 
97 
17 
16 


115 

97 


17 

16 


0639 
4061 


65787 
23435 


SECOND    CONTRACTED    METHOD. 

32.4687  )  790.755197 

53.42          141 3 

114 

17 

1 


1|42352 

ANALYSIS.  In  the  first  method 
of  contraction,  we  first  compare  the 
3  tens  of  the  divisor  with  the  79 
tens  of  the  dividend,  and  ascertain 
that  there  will  be  2  integral  places 
in  the  quotient ;  and  as  2  decimal 
places  are  required,  the  quotient 
must  contain  4  places  in  all.  Then 

assuming  the  four  left  hand  figures  of  the  divisor,  we  say  3246  is  con- 
tained in  7907,  2  times  ;  multiplying  the  assumed  part  of  the  divisor 
by  2.  and  carrying  2  units  from  the  rejected  part,  as  in  Contracted^ 
Multiplication  of  Decimals,  we  have  6494  for  the  product,  which  sub- 
tracted from  the  dividend,  leaves  1413  for  a  new  dividend.  Now, 
since  the  next  quotient  figure  will  be  of  an  order  next  below  the 
former,  we  reject  one  more  place  in  the  divisor,  and  divide  by  324, 
obtaining  4  for  a  quotient,  1299  for  a  product,  and  114  for  a  new  divi- 
dend. Continuing  this  process  till  all  the  figures  ot  the  divisor  are 


CONTRACTED  DIVISION.  135 

rejected,  we  have,  after  pointing  off  2  decimals  as  required,  24.35  for 
a  quotient.  Comparing  the  contracted  with  the  common  method,  we 
see  the  extent  of  the  abbreviation,  and  the  agreement  of  the  corres- 
ponding intermediate  results. 

In  the  second  method  of  contraction,  the  quotient  is  written  with 
its  first  figure  under  the  lowest  order  of  the  assumed  divisor,  and  the 
other  figures  at  the  left  in  the  reverse  order.  By  this  arrangement, 
the  several  products  are  conveniently  formed,  by  multiplying  each 
quotient  figure  by  the  figures  above  and  to  the  left  of  it  in  the  divisor, 
by  the  rule  for  contracted  multiplication,  (222),  and  the  remainders 
only  are  written  as  in  (112). 

22G.    From  these  illustrations  we  derive  the  following 
RULE.     I.  Compare  the  highest  or  left  hand  figure  of  the  divisor 
with  the  units  of  like  order  in  the  dividend,  and  determine  how 
many  figures  will  be  required  in  the  quotient. 

II.  For  the  first  contracted  divisor,  take  as  many  significant 
figures  from  the  left  of  the  given  divisor  as  there  are  places  re- 
'quired  in  the  quotient ;  and  at  each  subsequent  division  reject  one 
place  from  the  right  of  the  last  preceding  divisor. 

III.  In  multiplying  by  the  severed  quotient  figures,  carry  from 
the  rejected  figures  of  the  divisor  as  in  contracted  multiplication. 

NOTES.  —  1.  Supply  ciphers,  at  the  right  of  either  divisor  or  dividend,  when 
necessary,  before  commencing  the  work. 

2.  If  the  first  figure  of  the  quotient  is  written  under  tne  lowest  assumed  figure 
of  the  divisor,  and  the  other  figures  at  the  left  in  the  inverted  order,  the  several 
products  will  be  formed  with  the  greatest  convenience,  by  simply  multiplying 
each  quotient  figure  by  the  figures  above  and  to  the  left  of  it  in  the  divisor. 


EXAMPLES    FOR   PRACTICE. 

1.  Divide  27.3782  by  4.3267,  extending  the  quotient  to  3  deci- 
mal places.  Ans.  6.328  db. 

2.  Divide  487.24  by  1.003675,  extending  the  quotient  to  2 
decimal  places. 

3.  Divide  8.47326  by  75.43,  extending  the  quotient  to  5  deci- 
mal places. 

4.  Divide  .8487564  by  .075637,  extending  the  quotient  to  3 
decimal  places.  Ans.  11.221  dfc. 


13(3  DECIMALS. 

5.  Divide  478.325  by  1.43f ,  extending  the  quotient  to  3  deci- 
mal places.  Ans.  332.942  ±. 

6.  Divide  8972.436  by  756.3452,  extending  the  quotient  to  4 
decimal  places. 

7.  Divide  1  by  1.007633,  extending  the  quotient  to  6  decimal 
places.  Ans.  .992425  ±. 

8.  Find  the  quotient  of  .95372843  divided  by  44.736546,  true 
to  8  decimal  places. 

9.  Reduce  f  f  J-f  to  a  decimal  of  4  places.         Ans.  .7448  ±. 

CIRCULATING  DECIMALS. 

337.  Common  fractions  can  not  always  be  exactly  expressed  in 
the  decimal  form ;  for  in  some  instances  the  division  will  not  be 
exact  if  continued  indefinitely. 

338.  A  Finite  Decimal  is  a  decimal  which  extends  a  limited 
number  of  places  from  the  decimal  point. 

339.  An  Infinite  Decimal  is  a  decimal  which  extends   an* 
unlimited  number  of  places  from  the  decimal  point. 

330.  A  Circulating  Decimal  is  an  infinite  decimal  in  which 
a  figure  or  set  of  figures  is  continually  repeated  in  the  same  order ; 
as  .3333  +  ,  or  .437437437  +  . 

331.  A  Repetend  is  the  figure  or  set  of  figures   continually 
repeated.     When  a  repetend  consists  of  a  single  figure,  it  is  in- 
dicated by  a  point  placed  over  it;  when  it  consists  of  more  than 
one  figure,  a  point  is  placed  over  the  first,  and  one  over  the  last 
figure.    Thus,  the  circulating  decimals  .55555+  and  .324324324+, 
are  written,  5  and  .324. 

333.  A  repetend  is  said  to  be  expanded  when  its  figures  are 
continued  in  their  proper  order  any  number  of  places  toward  the 
right;  thus,  .24,  expanded  is  .2424+,  or  .242424242+. 

333.  Similar  Repetends  are  those  which  begin  at  the  same 
decimal  place  or  order;  as  .37  and  .5,  .24  and  .3/5, 1.56  and  24.3. 

334.  Conterminous  Repetends  are  those  which  end  at  the 
same  decimal  place  or  order;  as  .75  and  1.53,  .567,  and  3.245. 

NOTE. — Two  or  more  repetcnds  are  Similar  and  Conterminous  when  they  begt't* 
and  end  at  the  same  decimal  places  or  orders. 


CIRCULATING  DECIMALS.  137 

235.  A  Pure  Circulating  Decimal  is  one  which  contains  no 
figures  but  the  repetend;  as  .7,  or  .704. 

236.  A  Mixed  Circulating  Decimal  is  one  which  contains 
other  figures,  called  finite  places,  before  the  repetend;  as  .54,  or 
.013245,  in  which  .5  and  .01  are  the  finite  places. 

PROPERTIES    OF    FINITE    AND    CIRCULATING    DECIMALS. 

237.  The  operations  in  circulating  decimals  depend  upon  the 
following  properties. 

NOTE. — 1.  The  common  fractions  referred  to  are  understood  to  be  proper  frac- 
tions, in  their  lowest  terms. 

I.  Every  fraction  whose  denominator  contains  no  other  prime 
factor  than  2  or  5  will  give  rise  to  a  finite  decimal ;  and  the  num- 
ber of  decimal  places  will  be  equal  to  the  greatest  number  of  equal 
factors,  2  or  5,  in  the  denominator. 

For,  in  the  reduction,  every  cipher  annexed  to  the  numerator  mul- 
tiplies it  by  10,  or  introduces  the  two  prime  factors,  2  and  5,  and  also 
gives  1  decimal  place  in  the  result.  Hence  the  division  will  be  exact 
when  the  number  of  ciphers  annexed,  or  the  number  of  decimal 
places  obtained,  shall  be  equal  to  the  greatest  number  of  equal  factors, 
2  or  5,  to  be  canceled  from  the  denominater. 

II.  Every  fraction  whose  denominator  contains  any  other  prime 
factor  than  2  or  5,  will  give  rise  to  an  infinite  decimal. 

For,  annexing  ciphers  to  the  numerator  introduces  no  other  prime 
factors  than  2  and  5  ;  hence  the  numerator  will  never  contain  all  the 
prime  factors  of  the  denominator. 

III.  Every  infinite  decimal  derived  from  a  common  fraction  is 
also  a  circulating  decimal;    and  the    number    of  places    in    the 
repetend  must  be  less  than  the  number  of  units  in  the  denominator 
of  the  common  fraction. 

Eor,  in  every  division,  the  number  of  possible  remainders  is  limited 
to  the  number  of  units  in  the  divisor,  less  1 ;  thus,  in  dividing  by  7, 
the  only  possible  remainders  are  1,  2,  3,  4,  5,  and  6.  Hence,  in  the 
reduction  of  a  common  fraction  to  a  decimal,  some  of  the  remainders 
must  repeat  before  the  number  of  decimal  places  obtained  equals  the 
number  of  units  in  the  denominator ;  and  this  will  cause  the  inter- 
mediate quotient  figures  to  repeat. 
12* 


138  DECIMALS. 

NOTES. — 2.  It  will  be  found  that  the  number  of  places  in  the  repetend  is  always 
equal  to  the  denominator  less  1,  or  to  some  factor  of  this  number.  Thus,  the 
repetend  arising  from  ^  has  7  —  1  =  6  places  ;  the  repetend  arising  from  f  p  has 
i^l  =  5  places. 

3.  A  perfect  repetend  is  one  which  consists  of  as  many  places,  less  1,  as  there 
are  units  in  the  denominator  of  the  equivalent  fraction. 

4.  If  the  denominator  of  a  fraction  contains  neither  of  the  factors  2  and  5,  it 
will  give  rise  to  a  pure  repetend.     But  if  a  circulating  decimal  is  derived  from  a 
fraction  whose  denominator  contains  either  of  the  factors  2  or  5,  it  will  contain 
as  many  finite  places  as  the  greatest  number  of  equal  factors  2  or  5  in  the  de- 
nominator. 

IV.  If  to  any  number  we  annex  as  many  ciphers  as  there  are 
places  in  the  number,  or  more,  and  divide  the  result  by  as  many 
9's  as  the  number  of  ciphers  annexed,  both  the  quotient  and  re- 
mainder will  be  the  same  as  the  given  number. 

For,  if  we  take  any  number  of  two  places,  as  74,  and  annex  two 
ciphers,  the  result  divided  by  100  will  be  equal  to  74 ;  thus, 

7400  -~  100  =  74. 

Now  subtracting  1  from  the  divisor,  100,  will  add  as  many  units 
to  the  quotient,  74,  as  the  new  divisor,  99,  is  contained  times  in  74, 
(115,  II)  ;  thus, 

7400  -f-  99  =  74  +  Jf  or  74£f ; 

that  is,  if  two  ciphers  be  annexed  to  74,  and  the  result  be  divided  by 
99,  both  quotient  and  remainder  will  be  74.  In  like  manner,  annex- 
ing three  ciphers  to  74,  and  dividing  by  999,  we  have 

74000  -f-  999  =  74^ ; 
and  the  same  is  true  of  any  number  whatever. 

V.  Every  pure  circulating  decimal  is  equal  to  a  common  frac- 
tion whose  numerator  is  the  repeating  figure  or  figures,  and  whose 
denominator  is  as  many  9's  as  there  are  places  in  the  repetend. 

For,  if  we  take  any  fraction  whose  denominator  is  expressed  by 
some  number  of  9's,  as  f  $,  then  according  to  the  last  property,  annex- 
ing two  ciphers  to  the  numerator,  and  reducing  to  a  decimal,  we  have 
54  __  24{$.  In  like  manner,  carrying  the  decimal  two  places  farther, 
.24ff  =  .2424^  5  hence,  |£  =  24.  By  the  same  principle,  we  have 
£  =.2 ;  TV  =  -01 ;  ^  =  .02  ;  ifa  =  .OOi  ;  %%$  =  .324 ;  and  so  on.  And 
it  is  evident  that  all  possible  repetends  can  thus  be  derived  from  frac- 
tions whose  numerators  are  the  repeating  figures,  and  whose  denomi- 
nators are  as  many  9's  as  there  are  repeating  figures. 


CIRCULATING  DECIMALS. 

NOTE  5. — It  follows  from  the  last  property,  that  any  fraction  from  which  a  pure 
repetend  can  be  derived  is  reducible  to  a  form  in  which  the  denominator  is  some 
number  of  9's ;  thus  T8g-  =  §^f  n|  ',  &  =  »M-  This  is  true  of  eveiT  fraction, 
whose  denominator  terminates  with  1,  3,  7,  or  9. 

VI.  Any  repetend  may  be  reduced  to  another  equivalent  repe- 
tend, by  expanding  it,  and  moving  either  the  second  point,  or 
both  points,  to  the  right;  provided  that  in  the  result  they  be  so 
placed  as  to  include  the  same  number  of  places  as  are  contained 
in  the  given  repetend,  or  some  multiple  of  this  number. 

For,  in  every  such  reduction,  the  new  repetend  and  the  given  repe- 
tend, when  expanded  indefinitely,  will  give  results  which  are  identical. 
Thus,  .536  =  .536536,  or  .536536536,  or  .5365,  or  .53653,  or  .5365365, 
or  .53653653653  ;  because  each  of  these  new  repetends,  when  ex- 
panded, gives  .53653653653653653653+. 

NOTE  6.  —  If  in  any  reduction,  the  new  repetend  should  not  contain  the  samo 
number  of  places,  or  some  multiple  of  the  same  number,  as  the  given  repetend, 
we  should  not  have,  in  the  expansions,  the  same  figures  repeated  in  the  same 
order. 

REDUCTION. 
CASE    I. 

238.  To  reduce  a  pure  circulating  decimal  to  a 
common  fraction. 

1.  Reduce  .675  to  a  common  fraction. 

OPERATION.  ANALYSIS.     Since  the  repetend  has  3 

h*rk 675  25          places,  we  take  for  the  denominator  of 

*  — —   "n"a7j   —~*   "5 ? 

the  required   fraction   the   number  ex- 
pressed by  three  9's,  (237,    V).     Hence, 

RULE.  Omit  the  points  and  the  decimal  sign,  and  write  the 
figures  of  the  repetend  for  the  numerator  of  a  common  fraction, 
and  as  many  9's  as  there  are  places  in  the  repetend  for  the  de- 
nominator. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  .45  to  a  common  fraction.  Ans.  T5T. 

2.  Reduce  .66  to  a  common  fraction. 

3.  Reduce  .279  to  a  common  fraction.  Ans.         . 


140  DECIMALS. 

4.  Reduce  .423  to  a  common  fraction.  Ans.  T4T7T. 

5.  Reduce  .923076  to  a  common  fraction.  Ans.  jf  . 

6.  Reduce  .95121  to  a  common  fraction. 

7.  Reduce  4.72  to  a  mixed  number.  Ans.  4T8T. 

8.  Reduce  2.297  to  an  improper  fraction.  Ans.  f  f  . 

9.  Reduce  2.97  to  an  improper  fraction.  Ans.   l^°. 

NOTE.  —  According  to  237,  VI,  2.97  =  2.972. 

10.  Reduce  15.0  to  a  mixed  number.  Ans.  IS^I^. 

CASE    II. 

239.   To  reduce  a  mixed  circulating  decimal  to  a 
common  fraction. 

1.  Reduce  .0756  to  a  common  fraction. 

OPERATION.  ANALYSIS.  Since  .756  is  equal 

.0756  =  =  to       ,  .0756  will  be  -     of      «, 


2.  Reduce  .647  to  a  common  fraction. 

OPERATION.  ANALYSIS.    Reducing  the  finite 

04>r  __  _6_4     I      7  part  and  the  repetend  separately 

640  _  ^       7  to  fractions,  we  have  TVff  +  ^. 

=  —  —  —  ---  l~rr~7T          ^°   reduce   these    fractions  to  a 

common  denominator,   we   must 

_  640  —  64  -f  7  multiply  the  terms  of  the  first  by 

900  9  ;  but  the  numerator,  64,  may 

647  _  64  be  multiplied  by  9  by  annexing 

90Q  1  cipher  and  subtracting  64  from 

coo  ,,  I,      .  .       640  —  64     „ 

=  __  Ans.  r         '  glvmg       ()0o     ?     or 

the  first  fraction  reduced.     The 

^r^  numerator  of  the  sum  of  the  two 

.647  eiven  decimal.  fractions  will  therefore   be   640 

64   finite  figures.  -  64  +  7  =  583,  and  supplying 

>gq  the  common  denominator,  we  have 

||}g.      In  the   second  operation, 

__   Ans.  the  intermediate  steps  are  omitted. 

Hence  the  following 

RULE.     I.  From  the  given  circulating  decimal  subtract  the  finite 
j  and  the  remainder  will  be  the  required  numerator. 


CIRCULATING   DECIMALS. 

II.  Write  as  many  9's  as  there  are  figures  in  the  repetend,  with 
as  many  ciphers  annexed  as  there  are  finite  decimal  figures,  for 
the  required  denominator. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  .57  to  a  common  fraction.  Ans.  ||. 

2.  Reduce  .048  to  a  common  fraction.  Ans.  -£^. 

3.  Reduce  .6472  to  a  common  fraction. 

4.  Reduce  .6590  to  a  common  fraction.  Ans.  ||. 

5.  Reduce  .04648  to  a  common  fraction.  Ans.  -g423^. 

6.  Reduce  .1004  to  a  common  fraction. 

7.  Reduce  .9285714  to  a  common  fraction.  Ans.  l|. 

8.  Reduce  5.27  to  a  common  fraction.  Ans.  ff . 

9.  Reduce  7.0126  to  a  mixed  number.  Ans.  7^|g. 

10.  Reduce  1.58231707  to  an  improper  fraction.     Ans.  |£f . 

11.  Reduce  2.029268  to  an  improper  fraction. 

CASE   III. 

24O.  To  make  two  or  more  repetends  similar  and 
conterminous. 

1.  Make  .47,  .53675,  and  .37234  similar  and  conterminous. 

OPERATION.  ANALYSIS.     The  first  of 

.   .  .  the  given  repetends  begins 

.47        =  .47474747474747  -v  at  the  place  of  tenths?  the 

.53675  =  .53675675675675  V  Ans.      second  at  the  place  of  thou- 
.37234  =  .37234723472347  j  sandths,  and  the  third  at 

the  place  of  hundredths ; 

and  as  the  points  in  any  repetend  cannot  be  moved  to  the  left  over 
the  finite  places,  we  can  make  the  given  repetends  similar,  only  by 
moving  the  points  of  at  least  two  of  them  to  the  right. 

Again,  the  first  repetend  has  2  places,  the  second  3  places,  and  the 
third  4  places ;  and  the  number  of  places  in  the  new  repetends  must 
be  at  least  12,  which  is  the  least  common  multiple  of  2,  3,  and  4. 
We  therefore  expand  the  given  repetends,  and  place  the  first  point  in 
each  new  repetend  over  the  third  place  in  the  decimal,  and  the  second 
point  over  the  fourteenth,  and  thus  render  them  similar  and  conter- 
minous. Hence  the  following 


142  DECIMALS. 

RULE.  I.  Expand  the  repetends,  and  place  the  first  point  in 
each  over  the  same  order  in  the  decimal. 

II.  Place  the  second  point  so  that  each  new  repetend  shall  con- 
tain as  many  places  as  there  are  units  in  the  least  common  mul- 
tiple of  the  number  of  places  in  the  several  given  repetends. 

NOTK. — Since  none  of  the  points  can  be  carried  to  the  left,  some  of  them  must 
be  carried  to  the  right,  so  that  each  repetend  shall  have  at  least  as  many  finite 
places  as  the  greatest  number  in  any  of  the  given  repetends. 

EXAMPLES    FOR   PRACTICE. 

1.  Make  .43,  .57,  .4567,  and  .5037  similar  and  conterminous. 

2.  Make  .578,  .37,  .2485,  and  04  similar  and  conterminous. 

3.  Make  1.34,4.56,  and  .341  similar  and  conterminous. 

4.  Make  .5674,  .34,  .247,  and  .67  similar  and  conterminous. 

5.  Make  1.24,   .0578,   .4,   and  .4732147  similar  and   conter- 
minous. 

6.  Make  .7,  .4567,  .24,  and  .346789  similar  and  conterminous. 

7.  Make  .8,  .36,  .4857,  .34567,  and   .2784678943  similar  and 
conterminous. 

ADDITION  AND    SUBTRACTION. 

341.  The  processes  of  adding  and  subtracting  circulating  deci- 
mals depend  upon  the  following  properties  of  repetends : 

I.  If  two  or  more  repetends  are  similar  and  conterminous,  their 
denominators  will  consist  of  the  same   number  of  9's,  with  the 
same  number  of  ciphers  annexed.     Hence, 

II.  Similar  and  conterminous  repetends  have  the  same  denomi- 
nators and  consequently  the  same  fractional  unit. 

1.  Add  .54,  3.24  and,  2.785. 

OPERATION.  ANALYSIS.      Since    fractions   can   be 

54    =      54444  added  only  when  they  have  the  same 

3  9d          &9A9AV  fractional  unit,  we  first  make  the  repe- 

"^  . ""   I1  tends  of  the  given  decimals  similar  and 

2.785  as  2.78527  conterminous.    We  then  add  as  in  finite 

6.57214  decimals,  observing,  however,  that  the 

1  which  we  carry  from  the  left  hand 

column  of  the  repetends,  must  also  be  added  to  the  right  hand  column  ; 
for  this  would  be  required  if  the  repetends  were  further  expanded 
before  adding. 


CIRCULATING  DECIMALS  143 

2.  From  7.4  take  2.  7852. 

OPERATION.  ANALYSIS.     Since  one  fraction  can  be  subtracted 

7  4114.  ^rom  anotner  onty  when  they  have  the  same  frac- 

tional unit,  we  first  make  the  repetends  of  the  given 
decimals  similar  and  conterminous.     We  then  sub- 
4.6581  tract  as  in  finite  decimals ;  observing  that  if  both 

repetends  were  expanded,  the  next  figure  in  the 
subtrahend  would  be  8,  and  the  next  in  the  minuend  4  ;  and  the  sub- 
traction in  this  form  would  require  1  to  be  carried  to  the  2,  giving  1 
for  the  right  hand  figure  in  the  remainder. 

S5412.  From  these  principles  and  illustrations  we  derive  the 
following 

PwULE,  I.  When  necessary,  make  the  repetends  similar  and  con- 
terminous. 

II.  To  add , — Proceed  as  in  finite  decimals,  observing  to  increase 
the  sum  of  the  right  hand  column  by  as  many  units  as  are  carried 
from  the  left  hand  column  of  the  repetends. 

III.  To  subtract ;  —  Proceed  as  in  finite  decimals,  observing  to 
diminish  the  right  hand  figure  of  the  remainder  by  1,  when  the 
repetend  in  the   subtrahend  is   greater  than  the  repetend   of  the 
minuend. 

IV.  Place  the  points  in  the  result  directly  under  the  points  above. 

NOTS. — When  the  sum  or  difference  is  required  in  the  form  of  a  common  frac- 
tion, proceed  according  to  the  rule,  and  reduce  the  result. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  sum  of  2.4,  .32,  .567,  7.056,  and  4.37  ? 

Ans.  14.7695877. 

2.  What  is  the  sum  of  .478,  .321,  .78564,  .32,  .5,  and  .4326? 

Ans.  2.8961788070698. 

3.  From  .7854  subtract  .59.  Ans.  .1895258. 

4.  From  57.0587  subtract  27.31.  Ans.  29.7455. 

5.  What  is  the  sum  of  .5,  .32,  and  .12  ?  Ans.   1. 

6.  What  is  the  sum  of  .4387,  .863,  .21,  and  .3554  ? 

7.  What  is  the  sum  of  3.6537,  3.135,  2.564,  and  .53  ? 

8.  From  .432  subtract  .25.  Ans.  .18243. 

9.  From  7.24574  subtract  2.634.  Ans.  4.61. 


144 


DECIMALS. 


10.  From  .99  subtract  .433.  Ans.  .55656. 

11.  What  is  the  sum  of  4.638,  8.318,  .016,  .54,  and  .45? 

Aiis.  13|f. 

12.  From  .4  subtract  .23.  Ans.  ^. 

MULTIPLICATION   AND   DIVISION. 

S43.    1.  Multiply  2.428571  by  .063. 

OPERATION.  ANALYSIS.     We  first  re- 

duce  the  multiplicand  and 

.  .  "  7  multiplier  to  their  equiva- 

.063         =  Tyu  lent  fractions,  and  obtain 

y  X  T1TJ  =  T'/o  =  -154  Am.          V7  and  Tfr ;  then  V  X  T}T 

=  Tyo  =  .154. 

2.  Divide  .475  by  .3753. 

ANALYSIS.     The  dividend  re- 

OPERATION. 

.    .  duced  to  its  equivalent  common 

47^      =   475 

.    .  ~  fraction  is  $££,  and  the  divisor 

.61 oO  =  7j7j7j(j  reduced  to  its  equivalent  com- 

II!  X  If  38  =  L2^  -^s-          mon  ^action  is  fjft  ;   and  #$ 

H-  «*{f  =  -H  =  1-26. 

344.    From  these  illustrations  we  have  the  following 
RULE.     Reduce  the  given  numbers  to  common  fractions ;  then 
multiply  or  divide,  and  reduce  the  result  to  a  decimal. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  3.4  by  .72.  An$.  2.472. 

2.  Multiply  .0432  by  18.  Ans.  .7783. 

3.  Divide  .154  by  .2.  Ans.  .693. 

4.  Divide  4.5724  by  .7.  Ans.  5.878873601645. 

5.  Multiply  4.37  by  .27.  Ans.  1.182. 

6.  Divide  56.6  by  137.  Ans.  .41362530. 

7.  Divide  .428571  by  .54.  Ans.  .7857142. 

8.  Multiply  .714285  by  .27.  Ans.  .194805. 

9.  Multiply  3.456  by  .425.  Ans.  1.4710037. 

10.  Divide  9.17045  by  3.36.       Ans.  2.726350506748310. 

11.  Multiply  .24  by  .57.      Ans.  .1395775941230486685032. 


UNITED  STATES  MONEY. 


UNITED  STATES  MONEY. 

245.  By  Act  of  Congress  of  August  8,  1786,  the  dollar  was 
declared  to  be  the  unit  of  Federal  or  United  States  Money ;  and 
the  subdivisions  and  multiples  of  this  unit  and  their  denomina- 
tions, as  then  established,  are  as  shown  in  the 

TABLE. 

10  mills  make  1  cent. 

10  cents      "  1  dime. 

10  dimes     "  1  dollar. 

10  dollars  "  1  eagle. 

24G.    By  examining  this  table  we  find 

1st.  That  the  denominations  increase  and  decrease  in  a  tenfold 
ratio. 

2d.  That  the  dollar  being  the  unit,  dimes,  cents  and  mills  are 
respectively  tenths,  hundredths  and  thousandths  of  a  dollar. 

3d.  That  the  denominations  of  United  States  money  increase 
and  decrease  the  same  as  simple  numbers  and  decimals. 

Hence  we  conclude  that 

I.  United  States  money  may  be  expressed  according  to  tJie.  deci- 
mal system  of  notation. 

II.  United  States  money  may  be  added,  subtracted,  multiplied 
and  divided  in  the  same  manner  as  decimals. 

NOTATION    AND    NUMERATION. 

247.  The  character  $  before  any  number  indicates  that  it 
expresses  United  States  money.     Thus  $75  expresses  75  dollars. 

248.  Since  the  dollar  is  the  unit,  and  dimes,  cents  and  mills 
are  tenths,  hundredths  and  thousandths  of  a  dollar,  the  decimal 
point  or  separatrix  must  always  be  placed  before  dimes.     Hence, 
in  any  number  expressing  United  States  money,  the  first  figure  at 
the  right  of  the  decimal  point  is  dimes,  the  second  figure  is  cents, 
the  third   figure  is  mills,  and  if  there  are   others,  they  are  ten- 
thousandths,  hundred-thousandths,  etc.,  of  a  dollar.  Thus,  $8.3125 

13  K 


146  DECIMALS. 

expresses  8  dollars  3  dimes  1  cent  2  mills  and  5  tenths  of  a  mil] 
or  5  ten-thousandths  of  a  dollar. 

24:0.  The  denominations,  eagles  and  dimes,  are  not  regarded 
in  business  operations,  eagles  being  called  tens  of  dollars  and 
dimes  tens  of  cents.  Thus  $24.19  instead  of  being  read  2  eagles 
4  dollars  1  dime  9  cents,  is  read  24  dollars  19  cents.  Hence, 
practically,  the  table  of  United  States  money  is  as  follows : 

10  mills  make  1  cent. 
100  cents      "      1  dollar. 

25O.  Since  the  cents  in  an  expression  of  United  States  money 
may  be  any  number  from  1  to  99,  the  first  two  places  at  the  right 
of  the  decimal  point  are  always  assigned  to  cents.  Hence,  when 
the  number  of  cents  to  be  expressed  is  less  than  10,  a  cipher 
must  be  written  in  the  place  of  tenths  or  dimes.  Thus,  7  cents  is 
expressed  $.07. 

NOTES.  —  1.  The  half  cent  is  frequently  written  as  5  mills  and  vice  versd. 
Thus,  $.37*  =  $.375. 

2.  Business   men  frequently  write  cents   as   common   fractions   of  a  dollar. 
Thus,  $5.19  is  also  written  $5^^,  read  5  and  T'g9ff  dollars. 

3.  In  business  transactions,  when  the  final  result  of  a  computation  contains  5 
mills  or  more,  they  are  called  one  cent,  and  when  lens  than  5  they  are  rejected. 
Thus,  $2.198  would  be  called  $2.20,  and  $1.623  would  be  called  $1.62. 

EXAMPLES   FOR   PRACTICE. 

1.  Write  twenty-eight  dollars  thirty-six  cents. 

Am.  $28.36. 

2.  Write  four  dollars  seven  cents. 

3.  Write  ten  dollars  four  cents. 

4.  Write  sixteen  dollars  four  mills. 

5.  Write  thirty-one  and  one-half  cents. 

6.  Write  48  dollars  If  cents.  Arts.  $48  Olf. 

7.  Write  1000  dollars  1  cent  1  mill. 

8.  Write  3  eagles  2  dollars  5  dimes  8  cents  4  mills. 

9.  Write  6£  cents. 

10.  Head  the  following  numbers : 

$21.18  $10.01  $     .8125 

$164.05  $201.201  $15.08i 

$7.90  $5.37*  $96.005 


UNITED  STATES  MONEY. 


REDUCTION. 
Since  $1  =  100  cents  =  1000  mills,  it  is  evident, 

1st  That  dollars  may  be  changed  or  reduced  to  cents  by  an- 
nexing two  ciphers ;  and  to  mills  by  annexing  three  ciphers. 

2d.  That  cents  may  be  reduced  to  dollars  by  pointing  off  two 
figures  from  the  right;  and  mills  to  dollars  by  pointing  off  three 
figures  from  the  right. 

3d.   That  cents  may  be  reduced  to  mills  by  annexing  one  cipher. 

4th.  That  mills  may  be  reduced  to  cents  by  pointing  off  one 
figure  from  the  right. 

OPERATIONS    IN    UNITED    STATES    MONEY. 

S55J.  Since  United  States  Money  may  be  added,  subtracted, 
multiplied  and  divided  in  the  same  manner  as  decimals,  (24G, 
II),  it  is  evident  that  no  separate  rules  for  these  operations  are 
required. 

EXAMPLES    FOR    PRACTICE. 

1.  Paid  83475.50  for  building  a  house,  6310.20  for  painting, 
$1287. 37£  for  furniture,  and  $207. 12£  for  carpets;  how  much 
was  the  cost  of  the,  house  and  furniture  ?  Ans.  $5280.20. 

2.  Bought  a  pair  of  boots  for  $4.62J,  an  umbrella  for  SI. 75,  a 
pair  of  gloves  for  $.87 1,  a  cravat  for  $1,  and  some   collars   for 
$.62J;  how  much  was  the  cost  of  all  my  purchases? 

3.  Gave  $150  for  a  horse,  $175.84  for  a  carriage,  and  $62 £  for 
a  harness,  and   sold   the  whole  for  $390. 37  J;  how  much  did  I 
gain?  Ans.  $2.035. 

4.  A  man  bought  a  farm  for  $3800,  which  was  $190. 87£  less 
than  he  sold  it  for;  how  much  did  he  sell  it  for? 

5.  A  lady  bought  a  dress  for  $10|,  a  bonnet  for  $5£,  a  veil  for 
$2f,  a  pair  of  gloves  for  $.87*,  and  a  fan  for  $-|.      She  gave  the 
shopkeeper  a  fifty  dollar  bill;  how  much  money  should  he  return 
to  her?  '  Ans.  $29.875. 

G.  A  farmer  sold  150  bushels  of  oats  at  $.37 ?  a  bushel,  and  4 
cords  of  wood  at  $3 1  a  cord.  He  received  in  payment  84  pounds  of 


148  DECIMALS.  , 

sugar  at  61  cents  a  pound,  25  pounds  of  tea  at  $-f  a  pound,  2 
barrels  of  flour  at  $5.872,  and  the  remainder  in  cash;  how  much 
cash  did  he  receive?  Ana.  $39.125. 

7.  A  speculator  bought  264.5  acres  of  land  for  86726.     He 
afterward  sold  126.25  acres  for  $311  an  acre,  and  the  icmainder 
for  $33.75  an  acre;  how  much  did  he  gain  by  the  transaction  ? 

8.  A  merchant   going   to   New  York  to  purchase  goods,  had 
$11000.     He  bought  40  pieces  of  silk,  each  piece  containing  28 } 
yards,  at  $.80  a  yard;  300   pieces  of  calicoes,  with   an   average 
length  of  29  yards,  at  11  \  cents  a  yard  ;  20  pieces  of  broadcloths, 
each   containing  36.25   yards,  at  $3.875  a  yard;  112  pieces  of 
sheeting,   each   containing  30.5   yards,   at  $.06i   a  yard.     How 
much  had  he  left  with  which  to  finish  purchasing  his  stock  ? 

Ans.  SGiMiU/JA. 

9.  If  139  barrels  of  beef  cost    $2189.25,   how  much  will    1 
barrel  cost?  Ans.  $15.75. 

10.  If  396  pounds  of  hops  cost  $44.748,  how  much  are  they 
worth  per  pound  ?  Ans.  $.113. 

11.  Bought  10f  cords  of  wood  at  $44  a  cord,  and  received  for 
it  7.74  barrels  of  flour;  how  much  was  the  flour  worth  per  barrel  ? 

12.  If  a  hogshead  of  wine  cost  $287.4,  how  many  hogsheads 
can  be  bought  for  $4885.80  ?  Ans.  17. 

13.  A  butcher  bought  an  equal  number  of  calves  and  sheep  for 
$265;  for  the  calves  he  paid  $3;}  a  head,  and  for  the  sheep  $21 
a  head ;  how  many  did  he  buy  of  each  kind  ?  Ans.  40. 

14.  If  128   tons  of  iron  cost  $9632,  how   many   tons  can  be 
bought  for  $1730.75  ?  Ans.  23. 

15.  If  125  bushels  of  potatoes  cost  $41.25,  how  many  barrels, 
each  containing  2£  bushels,  can  be  bought  for  $112.20? 

16.  A  grocer  on  balancing  his  books  at  the  end  of  a  month, 
found  that  his  purchases  amounted  to  $2475.36,  and  his  sales  to 
SI 936. 40 ;  and  that  the  money  he  now  had  was  but  J  of  what  he 
had  at  the  beginning  of  the  month;  how  much  money  had  hu  at 
the  beginning  of  the  month?  Am*.  8131:7.40. 

17.  A  person  has  an  income  of  $3200  a  year,  and  his  expenses 
are  $138  a  month ;  how  much  can  he  save  in  8  years  ? 


UNITED  STATES  MONEY.  149 

18.  Sold  120  pieces  of  cloth  at  $45  f  a  piece,  and  gained  thereby 
$1026 ;  how  much  did  it  cost  by  the  piece  ?  Ans.  837.20. 

19.  A  flour  merchant  paid  $3088.25  for  some  flour.     He  sold 
425  barrels  at  $0}  a  barrel,  and  the  remainder  stood  him  in  $4.50 
a  barrel;  how  many  barrels  did  he  purchase  ?  Ans.  521. 

20.  If  36  engineers  receive  $6315.12  for  one  month's  work, 
how  many  engineers  will  $21927.50  pay  for  one  month  at  the  same 
rate?  Ans.  125. 

21.  A  person  having   $1378.56,  wishes  to  purchase  a  house 
worth  $2538,  and  still  have  $750  left  with  which  to  purchase  fur- 
niture; how  much  more  money  must  he  have?   Ans.  81900.44. 

22.  A  mechanic  earns  on  an  average  $1.87?  a  day,  and  works  22 
days  per  month.     If  his  necessary  expenses  are  $25:}  a  month, 
how  many  years  will  it  take  him  to  save  $1116,  there  being  12 
months  in  a  year?  Ans.  6  years 

23.  Bought  27.5  barrels  of  sugar  for  $453.75,  and  sold  it  at  a 
profit  of  $o.62  £  a  barrel;  at  what  price  per  barrel  was  it  sold  ? 

24.  A  man  expended  $70.15  in  the  purchase  of  rye  at  $.95  a 
bushel,  wheat  at  $1.37  a  bushel,  and  corn  at  $.73  a  bushel,  buying 
the  same  quantity  of  each  kind ;  how  many  bushels  in  all  did  he 

.  purchase  ?  Ans    69  bushels. 

25.  A  farmer  bought  a  piece  of  land  containing  375 \  acres,  at 
$22  i  per  acre,  and  sold  J  of  it  at  a  profit  of  $1032! ;  at  what 
price  per  acre  was  the  land  sold  ?  Ans.  $27.75. 

26    If  3}  cords  of  wood  cost  $11  37>},  how  much  will  20i  cords 
cost?  Ans.  S65.40f. 

27.  If  i  of  a  hundred   pounds   of  sugar  cost  86 1,  how  much 
can  be  bought  for  $46.75,  at  the  same  rate  ? 

Ans.  5.5  hundred  pounds. 

28.  A  man  sold  a 'wagon  for  $62.50,  and  received  in  payment 
12;y  yards  of  broadcloth  at  $3J  per  yard,  and  the  balance  in  coffee 
at  12;>  cents  per  pound;  how  many  pounds  of  coffee   did  he  re- 
ceive ?  Ans.  175  pounds. 

29.  Bought  320  bushels  of  barley  at  the  rate  of  16  bushels  for 
$10.04,  and  sold  it  at  the  rate  of  20  bushels  for  $17 £ ;  how  much 
was  my  profit  on  the  transaction  ?  Ans.  $79.20. 

13* 


150  DECIMALS. 

PROBLEMS 
INVOLVING  THE  RELATION  OF  PRICE,  COST,  AND  QUANTITY. 

PROBLEM  I. 

2o3.   Given,  the  price  and  the  quantity,  to  find  the  cost. 

ANALYSIS.  The  cost  of  3  units  must  be  3  times  the  price  of  1  unit ; 
of  8  units,  8  times  the  price  of  1  unit ;  of  f  of  a  unit,  f  times  the  price 
of  1  unit,  etc.  Hence, 

RULE.     Multiply  the  prize  of  ONE  by  the  quantity. 

PROBLEM    II. 

254.   Given,  the  cost  and  the  quantity,  to  find  the  price. 

ANALYSIS.  By  Problem  I,  the  cost  is  the  product  of  the  price  mul- 
tiplied by  the  quantity.  Now,  having  the  cost,  which  is  a  product, 
and  the  quantity,  which  is  one  of  two  factors,  we  have  the  product 
and  one  of  two  factors  given,  to  find  the  other  factor.  Hence, 

RULE.     Divide  the  cost  by  the  quantity. 

PROBLEM    III. 

25."».  Given,  the  price  and  the  cost,  to  find  the  quantity. 
ANALYSIS.     Reasoning  as  in  Problem  II,  we  find  that  the  cost  is 
the  product  of  two  factors,  and  the  price  is  one  of  the  factors,     Hence, 
RULE.     Divide  the  cost  Ly  the  price. 

PROBLEM    IV. 

2*>6.  Given,  the  quantity,  and  the  price  of  100  01 
1000,  to  find  the  cost. 

ANALYSIS.  If  the  price  of  100  units  be  multiplied  by  the  number 
of  units  in  a  given  quantity,  the  product  will  bo. 100  times  the  required 
result,  because  the  multiplier  used  is  100  times  the  true  multiplier. 
F<>r  a  similar  reason,  if  the  price  of  1000  units  be  multiplied  by  the 
number  of  units  in  a  given  quantity,  the  product  will  be  1000  times 
the  required  result.  These  errors  can  be  corrected  in  t\vo  ways, 

l.-t.   ]>y  dividing  the  product  by  100  or  1000,  as  the  case  may  bo;  or, 

2d.  By  reducing  the  given  quantity  to  hundreds  and  decimals  of  a 
hundred,  or  to  thousands  and  decimals  of  a  thousand.  Hence, 


PROBLEMS  IN  UNITED  STATES  MONEY. 

llULE.  Multiply  the  price  by  the  quantify  reduced  to  hundreds 
and  decimals  of  a  hundred,  or  to  thousands  and  decimals  of  a 
thousand. 

NOTE.  —  In  business  transactions  the  Roman  numerals  C  and  M  are  com- 
monly used  to  indicate  hundreds  and  thousands,  where  the  price  is  by  the  100 
or  1000, 

PROBLEM   V. 

257.  To  find  the  cost  of  articles  sold  by  the  ton  of 

2000  pounds. 

ANALYSIS.  If  the  price  of  1  ton  or  2000  pounds  be  divided  by  2, 
the  quotient  will  be  the  price  of  J  ion  or  1000  pounds.  We  then  have 
the  quantity  and  the  price  of  1000  to  find  the  cost.  Hence, 

HULE.  Divide  the  price  o/  1  ton  by  2,  and  multiply  the  quo- 
tient by  the  number  of  pounds  expressed  as  thousandths, 

EXAMPLES    IN    THE    PRECEDING    PROBLEMS. 

1.  What  cost  187  barrels  of  salt,  at  $1.32  a  barrel? 

^l/w.  $246.84. 

2.  What  cost  5  firkins  of  butter,  each  containing  70^  pounds, 
at  8T3ff  a  pound  ?  Ans.  $G6.09|. 

3.  If  the  board  of  a  family  be  $501.87^  for  1  year,  how  much 
is  it  per  day?  Ans.  $1.37|. 

4.  At  $  10  J  a  dozen,  how  many  dozen  of  eggs  can  be  bought 
for  $18.48?  Ana.  176^ 

5.  What  is  the  value  of  1  4U  sacks  of  ^uano,  each  sack  contain- 
ing 162^  pounds,  at  $17f  a  ton  ?  An*.  S201.906J. 

G.  What  will  be  the  cost  OA  3240  peach  trees  at  $16^  per  hun- 
dred? Ans.  $534.60. 

7.  At  $66.44  a  ton,  what  will  be  the  cost  of  842|  tons  of  rail- 
road iron?  Ans.  $55992.31. 

8.  A  gentleman  purchased  a  farm  of  325.5  acres  for  $10660^  ; 
how  much  did  it  cost  per  acre?  Ans.  $32.75. 

9.  What  will  be  the  cost  of  840  feet  of  plank  at  $1.94  per  C ; 
and  1262  pickets  at  $12^  per  M  ?  Ans.  $32.071. 

10.  At  $1^  a  bushel,  how  many  bushels  of  wheat  can  be  bought 
for  637.68|  ?  Ans.  25^  bushels. 


152  DECIMALS. 

11.  What  will  be  the  cost  of  2172  pounds  of  plaster,  at  $3.875 
a  ton?  Ans.  84.208f 

12.  What  cost  §  of  456  bushels  of  potatoes  at  $.37  J  a  bushel? 

13.  If  321  barrels  of  apples  cost  881.25,  what  is  the  price  per 
barrel?  Ans.  $2.50. 

14.  What  must  be  paid  for  24240  feet  of  timber  worth  $9.37 £ 
per  M.?  Ans.  $227*. 

15.  At  $5|  an  acre,  how  many  acres  of  land  can  be  bought  for 
$4234.37i?  Ans.  752J. 

16.  How  much  must  be  paid  for  972'  feet  of  boards  at  $20.25 
per  M,  1575  feet  of  scantling  at  $2.87|  per  C,  and  8756  feet  of 
lath  at  $7 }  per  M  ?  Ans.  $130.634}. 

17.  What  is  the  value  of  1046  pounds  of  beef  at  84  f  per  hun- 
dred pounds?  Ans.  $48.37f. 

18.  What  is  the  value  of  5840  pounds  of  anthracite  coal  at 
$4.7  a  ton,  and  4376  pounds  of  shamokin  coal  at  $5.25  a  ton? 

19.  At  $2.50  a  yard,  how  much  cloth  can  be  purchased  for  $2  ? 

20.  What  is  the  value  of  3700  cedar  rails  at  $5|  per  C  ? 

21.  How  much  is  the  freight  on  3840  pounds  from  New  York 
to  Baltimore,  at  $.96  per  100  pounds?  Ans.  $36.864. 

22.  What  is  the  value  of  9  pieces  of  broadcloth,  each  piece 
containing  271  yards,  worth  $21  a  yard  ?  Ans.  $715.87  J. 

23.  At  $.42  a  pound,  how  many  pounds  of  wool  may  be  bought 
for  $80.745?  Ans.  192}. 

24.  What  will  be  the  cost  of  327  feet  of  boards  at  $15£  per 
M;  672  feet  of  siding  at  $1.62J  per  C,  and  1108  bricks  at  $4} 
perM?  Ans.  $20.69|. 

25    At  $f  per  yard,  how  many  yards  of  silk  may  be  bought  for 
$15|?  Ans.   18. 

26.  How  much  must  be  paid  for  the  transportation  of  18962 
pounds  of  pork  from  Cincinnati  to  New  York,  at  $10  a  ton? 

27.  If  15J  yards  of  silk  cost  $27.9,  what  is  the  price  per  yard  ? 

28.  What  cost  27860  railroad  tics  at  $125.38  per  thousand  ? 

29.  If  .7  of  a  ton  of  hay  cost  $13|,  what  is  the  price  of  1  ton  ? 

30.  What  is  the  value  of  720  pounds  of  hay  at,  $12.7:")  .-i  ton, 
and  912  pounds  of  mill  feed  at  815 J  a  ton  ?         Ans.  $ll.(ir»S. 


ACCOUNTS  AND  BILLS. 


153 


LEDGER  ACCOUNTS. 

258.  A  Ledger  is  the  principal  book  of  accounts  kept  by  mer- 
chants and  accountants.  Into  it  are  brought  in  summary  form 
the  accounts  from  the  journal  or  day-book.  The  items  often  form 
long  columns,  and  accountants  in  adding  sometimes  add  more  than 
one  column  at  a  single  operation,  (G8). 


do 

42.17 

36.24 

18.42 

10.71 

194.30 

347.16 

40.00 

12.94 

86.73 

271.19 

103.07 

500.50 

7.59 

11.44 

81.92 

110.10 

107.09 

207.16 

97.20 

21.77 

150.15 

427.26 

31642 

114.64 

81.13 

37.50 


(20 

5  506.76 

19432 

427.90 

173.26 

71.32 

39.46 

152.60 

271.78 

320.00 

709.08 

48.50 

63.41 

56.00 

410.10 

372.22 

137.89 

276.44 

18.19 

27.96 

157.16 

94.57 

177.66 

327.40 

1132.16 

876.57 

179.84 


(30 

52371.67 

4571.84 

1690.50 

2037.09 

5094.46 

876.54 

679.81 

155.48 

4930.71 

3104.13 

1987.67 

5142.84 

27630 

522.71 

3114  60 

1776.82 

7152.91 

9328.42 

472.19 

321.42 

2423.79 

1600.81 

5976.27 

4318.19 

682,45 

3174.96 


$14763.84 

33276.90 

47061.39 

18242.76 

37364.96 

8410.31 

5724.27 

56317.66 

81742.73 

22431.27 

40163.55 

32189.60 

7063.21 

3451.09 

9200.00 

1807.36 

56768.72 

63024.27 

3618045 

90807.08 

28763.81 

37196.75 

4230  61 

3719.84 

1367.92 

8756.47 


ACCOUNTS  AND  BILLS. 

A  Debtor,  in  business  transactions,  is  a  purchaser,  c*  A 
person  who  receives  money,  goods,  or  services  from  another;  and 
20O.    A  Creditor  is  a  seller,  or  a  person  who  parts  with 
money,  goods,  or  services  to  another. 


154  DECIMALS. 

961.    An  Account  is  a  registry  of  debts  and  credits. 

NOTES. — 1.  An  account  should  always  contain  the  nnmes  of  both  the  parties  to 
the  transaction,  the  price  or  value  of  each  item  or  article,  and  the  date  of  the 
transaction. 

2.  Accounts  may  have  only  one  side,  which  may  be  either  debt  or  credit;  or  it 
may  have  two  sides,  debt  and  credit. 

969.  The  Balance  of  an  Account  is  the  difference  between 
the  amount  of  the  debit  and  credit  sides.  If  the  account  have 
only  one  side,  the  balance  is  the  amount  of  that  side. 

963.  An  Account  Current  is  a  full  copy  of  an  account, 
giving  each  item  of  both  debit  and  credit  sides  to  date. 

96-4.  A  Bill,  in  business  transactions,  is  an  account  of  money 
paid,  of  goods  sold  or  delivered,  or  of  services  rendered,  with  the 
price  or  value  annexed  to  each  item. 

96*5,  The  Footing  of  a  Bill  is  the  total  amount  or  cost  of  all 
the  items. 

NOTE. — A  bill  of  goods  bought  or  sold,  or  of  services  received  or  rendered  at 
a  single  transaction,  and  containing  only  one  date,  is  often  called  a  BUI  of  Par- 
cels ;  and  an  account  current  having  only  one  side  is  sometimes  called  a  Bill 
of  Items. 

966.  In  accounts  and  bills  the  following  abbreviations  are  in 
general  use : 

Dr.  for  debit  or  debtor ; 
Cr.  for  credit  or  creditor; 
a|c.  or  acc't  for  account; 

@  for  at  or  by ;  when  this  abbreviation  is  used  it  is  always 
followed  by  the  price  of  a  unit.  Thus,  3  yd.  cloth  @  $1.25,  sig- 
nifies 3  yards  of  cloth  at  $1.25  per  yard;  £  lb.  tea  @  $.75,  signi- 
fies J  pound  of  tea  at  $.75  per  pound. 

967.  When  an  account  current  or  a  bill  is  settled  or  paid, 
the  fact  should  be  entered  on  the  same  and  signed  by  the  creditor, 
or  by  the  person  acting  for  him.     The  *|e.  or  bill  is  then  said  to 
be  rrrrijttrtf.     Accounts  and  bills  may  be  settled,  balanced  and 
receipted  by  the  parties  to  the  same,  or  by  agents,  clerks  or  attor- 
neys authorized  to  transact  business  for  the  parties. 


ACCOUNTS  AND  BILLS.  155 

EXAMPLES    FOR    PRACTICE. 

Required,  the  footings  and  balances  of  the  following  bills  and 
accounts. 

(i.) 

BiU:  receipted  by  clerk  or  agent. 

NEW  YORK,  July  10,  1860. 
Mr.  JOHN  C.  SMITH, 

Bo't  of  HILL,  GROVES  &  Co., 
10  yd.  Cassimere,  @  $2.85 

16    «    Blk.  Silk,  «      1.12J 

72    «    Ticking,  "         .14 

42    «    Bid.  Shirting, 
12    «    Pressed  Flannel, 
24£  "    Scotch  Plaid  Prints, 

Rec'd  Payment, 

HILL,  GROVES  &  Co., 

By  J.  W.  HOPKINS. 
(20 
Bill :  receipted  by  the  selling  party. 

CHICAGO,  Sept.  20,  1861. 
CHASE  &  KENNARD, 

Bot  of  McDou0AL,  FENTON  &  Co., 
125    pr.    Boys'  Thick  Boots, 
275     «        "      Calf        « 
180     «        "     Kip        « 
210     «        .«      Brogans, 
80     "     Women's  Fox'd  Gaiters,  " 
95     "  "         Opera    Boots, 

175     "  "         Enameled  " 

8  cases  Men's  Calf  Boots, 
3     "     Congress  Pump  Boots, 
1     "     Drill,  958  yd., 
40  gross  Silk  Buttons, 

Rec'd  Payment, 

McDouGAL,  FENTON  &  Co. 


156  DECIMALS. 

(3.) 

BUI :  settled  by  note. 

NEW  YORK,  May  4,  1860. 
SMITH  &  PERKINS, 

Bo't  of  KENT,  LOWBER  &  Co., 
40  chests  Green  Tea,  @  $27.50 


25  "  Black  " 
16  "  Imperial  " 
12  sacks  Java  Coffee, 


19.20 
48.10 
17.75 


20  bbl.      Coffee  Sugar,  (A)    «  26.30 

«  31.85 

«  4.12J 

"  2.90 


15     "       Crushed    " 
36  boxes  Lemons, 
42     "       Oranges, 
25     "       Eaisins, 


Rec'd  Payment,  l>y  note  at  6  mo. 


(4-) 


$3951.00 


KENT,  LOWBER  &  Co. 


Bill :  paid  by  draft,  and  receipted  by  Cleric. 

NEW  ORLEANS,  April  28,  1861. 

JAMES  CARLTON  &  Co. 

Bo't  of  WILLARD  &  HALE. 

150  bbl.  Canada  Flour,  @  $6.25 

275    "  Genesee    " 

170    "  Philada.    u 

326  bu.  Wheat, 

214    «  Corn, 

300    "  Barley, 

500    "  Rye, 


7.16 

5.87J 

1.621 

.82 

.91 

1.06 


$5413.48 


Rec'd  Payment,  by  Draft  on  N.  Y. 


R.  S.  CLARKE, 
For  WILLARD  &  HALE. 


ACCOUNTS  AND  BILLS. 


157 


(5.) 
Account  Current  ;  not  balanced  or  settled. 

PHILADELPHIA,  Nov.  1,  1860. 
MR.  JAMES  CORNWALL, 

To  DODGE  &  SON,  Dr. 

April  15,  To  24  tons  Swedes  Iron,            @  $64.30     $ 

«      "  "  15  cwt.  Eng.  Blister  Steel,    "      10.25 

June  21,  «  7  doz.  Hoes,  (Trowel  Steel)  " 

Aug.  10,  "  25    "    Buckeye  Plows,         " 

Oct.      3,  "  14  Cross-cut  Saws,                 " 

"       "  "  27  cwt.  Bar  Lead,                  " 

"       "  "  1840  Ibs.  Chain,                     " 


7.78 

8.45 

16.12* 

5.90 


May  25, 

July  14, 

tt      tt 

Sept.   5, 

it      10 


By  20  M.  Boards, 


Or.. 

@  $17.60 


50  M.  Shingles,    "  3.12* 

42  M.  Plank,        "  9.37 £ 
Draft  on  New  York, 

75  C.  Timber,     @  3.10 

36  C.  Scantling,  "  .871 


$1000 


Dr. 


Bal.  Due  DODGE  &  SON,        §35(5.51 


(6.) 

Account  Current,  another  form  ;  balanced  ~by  note. 
WM.  RICHMOND  &  Co.  in  a|c.  current  with  WOOD  &  POWELL. 


Or. 


I860 

18  tO 

July 

Aug. 

2 

17 

To  896  pounds  butter,       $.23 
'•  872       "        cheese,          .09^ 

Nov. 

JJ 

By  61  barrels  apple?,  $2.25 
4i  70  bushels  turnips,  .22 

!4 

"  481l<    «        lard,             .ll3^ 

Dec. 

1 

"  56     <•  dried  apples,  .SlU 

Oct. 

4 

"  509-%    "       tallow,        J8U 

•• 

J-. 

"  31  drums  figs,           -68% 

« 

'8 
31 

"  81  dozen  esss,                 -1% 
"  15  barrels  salt,              1.40 

1861 
Jan. 

•2 

'•'  Note  at  3  mo.  to  Bal. 

Dec. 

15 

"  41  hams,  96S%  pounds,  .12j^ 

-==:S;=:^^ 

1                                                          1 

566 

2fi 

^""a=^^ 

BOSTON,  Jan.  1, 1861. 


WOOD  &  POWBXL. 


158  DECIMALS,  ', 

PROMISCUOUS   EXAMPLES. 

1.  What  cost  12|  cords  of  wood®  S4.87H   Ans.  $61.54+. 

2.  At  $.371  per  bushel,  how  many  barrels  of  potatoes,  each 
containing  2J  bushels,  can  be  purchased  for  $33.75?     Ans.  36. 

3.  If  36  boxes  of  raisins,  each  containing  36  pounds,  can  be 
bought  for  $97.20,  what  is  the  price  per  pound  ?      Ans.  $.075. 

4.  If  .625  of  a  barrel  of  flour  be  worth  $5.35,  what  is  a  barrel 
worth  ?  Ans.  $8.56. 

5.  What  is  the  difference  between  |  of  a  hundredth,  and  -i  of 
a  tenth  ?  Ans.  .025. 

6.  What  is  the  product  of  8143909jj  X  26-J-f  correct  to  2  decimal 
places  ? 

7.  A  drover  bought  5  head  of  cattle  @  $75,  and  12  head  @ 
$68 ;  at  what  price  per  head  must  he  sell  them  to  gain  $118  on 
the  whole?  Ans.  $77. 

8.  If  1  pound  of  tea  be  worth  $.62£,  what  is  .8  of  a  pound 
worth?  Ans.  $.5. 

9.  A  person   having  $27.96,  was  desirous  of  purchasing  an 
equal  number  of  pounds  of  tea,  coffee,  and  sugar;   the  tea  @ 
$.872-,  the  coffee  @  $.18f,  and  the  sugar  @  $.10J.     How  many 
pounds  of  each  could  he  buy?  Ans.  24. 

10.  If  a  man  travel  13543.47  miles  in  365i  days,  how  far  does 
he  travel  in  |  of  a  day  ?  Ans.  32.445  miles. 

11.  Bought  100  barrels  of  flour  @  $5.12£,  and  250  bushels  of 
wheat  @  $1.06i.     Having  sold  75  barrels  of  the  flour  @  $6£, 
and  all  the  wheat  @  $lf,  at  what  price  per  barrel  must  the  re- 
mainder of  the  flour  be  sold,  to  gain  $221. 87£  on  the  whole  invest- 
ment? Ans.  $6.75. 

12.  If  114.45  acres  of  land  produce  4580.289  bushels  of  pota- 
toes, how  many  acres  will  be  required  to  produce  120.06  bushels? 

Ans.  3. 

13.  Divide  .0172JJ  by  .03-^,  and  obtain  a  quotient  true  to  4. 
decimal  places.  Ans.  .5625. 

14.  Divide  13.5  by  21  hundredths.  Ans.  600. 

15.  A  man  agreed  to  build  59.5  rods  of  wall;  having  built  8.5 


PROMISCUOUS  EXAMPLES.  159 

rods  in  5  days,  how  many  days  will  be  required  to  finish  the  wall 
at  the  same  rate?  Ans.  30  days. 

16.  A  farmer  exchanged  28£  bushels  of  oats  worth  $.37*  per 
bushel,  and  453  pounds  of  mill  feed  worth  $.75  per  hundred,  for 
12520  pounds  of  plaster;   how  much  was  the  plaster  worth  per 
ton  ?  Ans.  $2.25. 

17.  A  farmer  sold  to  a  merchant  3  loads  of  hay  weighing  re- 
spectively 1826,  1478,  and  1921  pounds,  at  $8.80  per  ton,  and 
281  pounds  of  pork  at  $5.25  per  hundred.  He  received  in  exchange 
31  yards  of  sheeting  @  $.09,  6*  yards  of  cloth  @  $4.50,  and  the 
balance  in  money;  how  much  money  did  he  receive? 

18.  If  35  yards  of  cloth  cost  $122.50,  what  will  be  the  cost  of 
29  yards?  Ana.  $101.50. 

19.  A  speculator  bought  1200  bushels  of  corn  @  8.56}.     He 
sold  375 o   bushels  @  $.60.   At  what  price  must  he  sell  the  re- 
mainder, to  gain  $168.675  on  the  whole? 

20.  If  a  load  of  plaster  weighing  1680  pounds  cost  $2.856,  how 
much  will  a  ton  of  2000  pounds  cost?  Ans.  $3.40. 

21.  If  .125  of  an  acre  of  land  is  worth  $15|,  how  much  are 
25.42  acres  worth  ? 

22.  A  farmer  had  150  acres  of  land,  which  he  could  have  sold 
at  one  time  for  $100  an  acre,  and  thereby  have  gained  $3900;  but 
after  keeping  it  for  some  time  he  was  obliged  to  sell  it  at  a  loss 
of  $2250.     How  much  an  acre  did  the  land  cost  him,  and  how 
much  an  acre  did  he  sell  it  for? 

23.  A  lumber  dealer  bought  212500  feet  of  lumber  at  $14.375 
per  M,  and  retailed  it  out  at  $1.75  per  C;  how  much  was  his 
whole  gain  ? 

24.  If  10  acres  of  land  can  be  bought  for  $545,  how  many 
acres  can  be  bought  for  $17712.50  ?  Ans.  325. 

25.  How  much  is  the  half  of  the  fourth  part  of  7  times  224.56  ? 

Ans.  196.49. 

26.  Sold    10450    feet   of  timber  for   $169.8125,  and    gained 
thereby  $39.18| ;  how  much  did  it  cost  per  C  ?        Ans.  $1.25. 

27.  If  $6.975  be  paid  for  .93  of  a  hundred  pounds  of  pork, 
how  much  will  1  hundred  pounds  cost  ? 


160  DECIMALS. 

28.  Three  hundred  seventy-five  thousandths  of  a  lot  of  dry 
goods,  valued  at  $4000,  was  destroyed  by  fire ;  how  much  would 
a  firm  lose  who  owned  .12  of  the  entire  lot [  Ans.  8180. 

29.  Reduce  (77  -^-Jf)  X  |  of  1  to  a  decimal.         Ans.  .15. 

30.  If  7.5  tons  of  hay  are  worth  375  bushels  of  potatoes,  and 
1  bushel  of  potatoes  is  worth  $.33|,  how  much  is  1   ton  of  hay 
worth?  Ans.  $16.66§. 

31.  A  person  invested  a  certain  sum  of  money  in  trade;  at  the 
end  of  5  years  he  had  gained  a  sum  equal  to  84  hundredths  of  it, 
and  in  5  years  more  he  had  doubled  this  entire  amount.     How 
many  times  the  sum  first  invested  had  he  at  the  end  of  the  10 
years?  Ans.  3.68  times. 

32.  A  miller  paid  $54  for  grain,  T30  of  it  being  barley  at  $.62* 
per  bushel,  and  |  of  it  wheat  at  $1.87*  per  bushel;  the  balance 
of  the  money,  he  expended  for  oats  at  $.37 *  per  bushel.     How 
many  bushels  of  grain  did  he  purchase  ?  Ans.  40. 

33.  A  merchant  tailor  bought  27  pieces  of  broadcloth,  each 
piece  containing  19*  yards,  at  $4.31|  a  yard;  and  sold  it  so  as  to 
gain  $381.87*,  after  deducting  $9.62*  for  freight.     How  much 
was  the  cloth  sold  for  per  yard  ?  Ans.  $5.06?. 

34.  Bought  1356  bushels  of  wheat  @  Sl.lSf ,  and  736  bushels 
of  oats  @  $.41 ;  I  had  870  bushels  of  the  wheat  floured,  and  dis- 
posed of  it  at  a  profit  of  8235. 87 *,  and  I  sold  528  bushels  of  tho 
oats  at  a  loss  of  $13.62*.     I  afterward  sold  the  remainder  of  the 
wheat  at  $1.12*  per  bushel,  and  the  remainder  of  the  oats  at  $.31 
per  bushel ;  did  I  gain  or  lose,  and  how  much  ? 

Ans.  I  gained  $171.07*. 

35.  The  sum  of  two  fractions  is  |||,  and  tUeir  difference  is 
\ }}  1 ;  what  are  the  fractions  ? 

36.  A  manufacturer  carried  on  business  for  3  years.     The  first 
year  he  gained  a  sum  equal  to  ^  of  his  original  capital;  the  second 
yr-ar  he  lost  jL  of  what  he  had  at  the  end  of  the  first  year;  the 
third  year  he  gained  |  of  what  he  had  at  the  end  of  th»>  second 
year,  and  he  then  had  $28585.70.     How  much  had  lu-  gained  in 
the  3  years?  Ans.  $10594.70 


CONTINUED  FRACTIONS.  161 


CONTINUED  FRACTIONS. 

968.  If  we  take  any  fraction  in  its  lowest  terms,  as  if,  and 
divide  both  terms  by  the  numerator,  we  shall  obtain  a  complex 
fraction,  thus : 

13       1 


13 

Reducing  ^,  the  fractional  part  of  the  denominator,  in  the  same 

manner,  we  have, 

13       1 

54  ~~  4  4-  1 

2~ 

Expressions  in  this  form  are  called  continued  fractions.     Hence, 

269.  A  Continued  Fraction  is  a  fraction  whose  numerator  is 
1,  and   whose  denominator  is   a  whole   number  plus   a  fraction 
whose  numerator  is  also  1,  and  whose  denominator  is  a  similar 
fraction,  and  so  on. 

270.  The  Terms  of  a  continued  fraction  are  the  several  sim- 
ple  fractions   which 'form  the   parts   of  the   continued   fraction. 
Thus,  the  terms  of  the  continued  fraction  given  above  are,  4,  &, 
and  £. 

CASE  i. 

271.  To  reduce  any  fraction  to  a  continued  fraction. 

1.  Reduce  ^||  to  a  continued  fraction. 

OPERATION.  ANALYSIS.    We  divide  the  denominator, 

109  _  1  339,  by  the  numerator,  109,  and  obtain  3 

339        3  -f  1  for  the  denominator  of  the  first  term  of 

9    i    1         the  continued  fraction.    Then  in  the  same 
j-^        manner  we  divide  the  last  divisor,  109,  by 
•  the  remainder,  12,  and  obtain  9  for  the  de- 

nominator of  the  second  term  of  the  continued  fraction.  In  like  man- 
ner we  obtain  12  for  the  denominator  of  the  final  term.  Hence  the 
following 

RULE.      I.  Divide   the  greater   term   by  the   less,  and  the  last 
divisor  by  the  last  remainder,  and  so  on,  till  there  is  no  remainder. 
14*  L 


162  CONTINUED  FRACTIONS.  \ 

II.  Write  1  for  the  numerator  of  each  term  of  the  continued 
fraction,  and  the  quotients  in  their  order  for  the  respective  denom- 
inators. 

.     •          EXAMPLES    FOR   PRACTICE. 

1.  Reduce  -fiff^  to  a  continued  fraction. 

Ans.  L_ 


2.  Reduce  Jf4?  to  a  continued  fraction. 

3.  Reduce  ^?Mnf  to  a  continued  fraction. 

o  j  o  y  u  i 

4.  Reduce  -f-^  to  a  continued  fraction. 

CASE    II. 

272.  To  find  the  several  approximate  values  of  a 
continued  fraction. 

An  Approximate  Value  of  a  continued  fraction  is  the  simple 
fraction  obtained  by  reducing  one,  two,  three,  or  more  terms  of  the 
continued  fraction. 

273.  1.  Reduce   T3g85   to   a  continued   fraction,  and  find  its 
approximate  values. 

OPERATION. 

38        1 

—  =  —.  —  ,  the  continued  fraction. 
lui 


2  +  1 
5 


—  1,  1st  approx.  value. 


X  2  +  1  _    3X2  +  1  _  J  ,  3d 
13  X  2  +  4       30' 


-          4th 
"        ' 


_ 

30X6+13"  103' 


CONTINUED  FRACTIONS. 

ANALYSIS.  We  take  \,  the  first  term  of  the  continued  fraction, 
for  the  first  approximate  value.  Reducing  the  complex  fraction 
formed  by  the  first  two  terms  of  the  continued  fraction,  we  have  T35 
for  the  second  approximate  value.  In  like  manner,  reducing  the  first 
three  terms,  we  have  ^  for  the  third  approximate  value.  By  exam-- 
ining  this  last  process,  we  perceive  that  the  third  approximate  value, 
3Tff,  is  obtained  by  multiplying  the  terms  of  the  preceding  approxima- 
tion, -pj,  by  the  denominator  of  the  third  term  of  the  continued  frac- 
tion, 2,  and  adding  the  corresponding  terms  of  the  first  approximate 
value.  Taking  advantage  of  this  principle,  we  multiply  the  terms  of 
S7iy  by  the  4th  denominator,  5,  in  the  continued  fraction,  and  adding 
the  corresponding  terms  of  T:  ^,  obtain  ^y^,  the  4th  approximate  value, 
which  is  the  same  as  the  original  fraction.  Hence  the  following 

RULE.  I.  For  the  first  approximate  value,  take  the  first  term 
of  the  continued  fraction. 

II.  for  the  second  approximate  value,  reduce  the  complex  frac- 
tion formed  ly  the  first  two  terms  of  the  continued  fraction. 

III.  For  each  succeeding  approximate  value,  multiply  bofh  nu- 
merator and  denominator  of  the  last  preceding  approximation  l)ij 
the  next  denominator  in  the  continued  fraction,  and  add  to  the  cor- 
responding products  respectively  the  numerator  and  denominator  of 
the  preceding  approximation. 

NOTES.  —  1.  When  the  given  fraction  is  improper,  invert  it,  and  reduce  this 
result  to  a  continued  fraction;  then  invert  the  approximate  values  obtained 
therefrom. 

2.  In  a  series  of  approximate  values,  the  1st,  3d,  5th,  etc.,  are  greater  than 
the  given  fraction  ;  and  the  2il,  4th,  Oth,  etc.,  are  less  than  the  given  fraction. 

EXAMPLES    FOR    PRACTICE. 

1.  Find  the  approximate  values  of  r6/F. 

^«-  a.  i.  if-  T%- 

2.  Find  the  approximate  values  of  ^\. 

A,,-       1        4         5         39         83 
A.T18.     4,   T7,   TTJ,   T64-,    34^. 

3.  What  are  the  first  three  approximate  values  of 

4  n  i       1        5 
All*.     TJ,    gg 

4.  What  are  the  first  five  approximate  values  of  f  §  J  ? 


5.  Reduce  |£  to  the  form  of  a  continued  fraction,  and  find  the 
value  of  each  approximating  fraction. 


164  COMPOUND  NUMBERS. 


COMPOUND  NUMBEKS. 

.  A  Compound  Number  is  a  concrete  number  expressed 
in  two  or  more  denominations,  (1O). 

27o.  A  Denominate  Fraction  is  a  concrete  fraction  whose 
integral  unit  is  one  of  a  denomination  of  some  compound  number. 
Thus,  |  of  a  day  is  a  denominate  fraction,  the  integral  unit  being 
one  day;  so  are  |  of  a  bushel,  |  of  a  mile,  etc.,  denominate  irac- 
tions. 

27G.  In  simple  numbers  and  decimals  the  scale  is  uniform, 
and  the  law  of  increase  and  decrease  is  by  10.  But  in  compound 
numbers  the  scale  of  increase  and  decrease  from  one  denomination 
to  another  is  varying,  as  will  be  seen  in  the  Tables. 

MEASURES. 

97T.  Measure  is  that  by  which  extent,  dimension,  capacity 
or  amount  is  ascertained,  determined  according  to  some  fixed 
standard. 


The  process  by  which  the  extent,  dimension,  capacity,  or  amount  is 
ascertained,  is  called  Measuring;  and  consists  in  comparing  the  thing  to  bo 
measured  with  some  conventional  standard. 

Measures  are  of  seven  kinds  : 

1.  Length.  4.  Weight,  or  Force  of  Gravity. 

2.  Surface  or  Area.  5.   Time. 

3.  Solidity  or  Capacity.         6.   Angles. 

7.   Money  or  Value. 

The  first  three  kinds  may  be  properly  divided  into  two  classes  —  • 
Measures  of  Extension,  and  Measures  of  Capacity. 

MEASURES  OF  EXTENSION. 

278.   Extension  has  three  dimensions  —  length,  breadth,  and 
thickness. 

A  Line  has  only  one  dimension  —  length. 

A  Surface  or  Area  has  two  dimensions  —  length  and  breadth. 


MEASURES  OF  EXTENSION. 

A  Solid  or  Body  has  three  dimensions  —  length,  breadth,  and 
thickness. 

I.  LINEAR  MEASURE. 

S79.  Linear  Measure,  also  called  Long  Measure,  is  used  in 
measuring  lines  or  distances. 

The  unit  of  linear  measure  is  the  yard,  and  the  table  is  made 
up  of  the  divisors,  (feet  and  inches,)  and  the  multiples,  (rods, 
furlongs,  and  miles,)  of  this  unit. 

TABLE. 

12  inches  (in.)  make  1  foot, ft. 

3  feet  "  1  yard, yd. 

5J  yards,  or  16  J  feet,  "  1  rod, rd. 

820  rods  "      1  statute  mile,,  .mi. 

UNIT   EQUIVALENTS. 

ft.  in. 

yd.                       1  =  12 

rd.                      1      =             3  =  36 

mi                 1     =           5i  =         16-i-  =  198 

1     =     320     =     1760     =     5280  =  63360 

SCALE— ascending,  12,  3,  5£,  40,  8 ;  descending,  8,  40,  5£,  3,  12. 
The  following  denominations  are  also  in  use  :  — 


3       barleycorns  make  1  inch, 


inches  "      1  hand, 


used  by  shoemakers  in  measuring 
the  length  of  the  foot. 


used  in  measuring  the  height  of 
horses  directly  over  the  fore  feet. 

9  "  "1  span. 

21.888    "  "      1  sacred  cubit. 

3       feet  "      1  pace. 

6         "  "      1  fathom,  used  in  measuring  depths  at  sea. 

X.152J  statute  mi.    ••      1  geographic  mile,     **"**  ** 


3       geographic  "    "      1  league. 


60  "  or  |    -•    I  f  of  latitude  on  a  meridian  or  of 

69.16  statute         "    "  j  °         (  longitude  on  the  equator. 

360       degrees  "      the  circumference  of  the  earth. 

NOTES.  —  1.  For  the  purpose  of  measuring  cloth  and  other  goods  sold  by  the 
yard,  the  yard  is  divided  into  halves,  fourths,  eighths,  and  sixteenths.  The  old 
table  of  cloth  measure  is  practically  obsolete. 

2.  A  span  is  the  distance  that  can  be  reached,  spanned,  or  measured  between 
the  end  of.  the  middle  finger  and  the  end  of  the  thumb.     Among  sailors  8  spans 
are  equal  to  1  fathom. 

3.  The  geographic  mile  is  ^  of  3^  or  7,  j-'g^Ti  of  the  distance  round  the  center 
of  the  earth.     It  is  a  small  fraction  more  than  1.15  statute  miles. 


166 


COMPOUND    NUMBERS. 


4.  The  length  of  a  degree  of  latitude  varies,  being  68.72  miles  at  the  equator, 
68.9  to  69.05  miles  in  middle  latitudes,  and  69.30  to  69.34  miles  in  the  polar 
regions.  The  mean  or  average  length,  as  stated  in  the  table,  is  the  standard 
recently  adopted  by  the  U.  S.  Coast  Survey.  A  degree  of  longitude  is  greatest 
at  the'  equator,  where  it  is  69.16  miles,  and  it  gradually  decreases  toward  the 
poles,  where  it  is  0. 

SURVEYORS'  LINEAR  MEASURE. 

38O.  A  Ghinter's  Chain,  used  by  land  surveyors,  is  4  rods 
.or  66  feet  long,  and  consists  of  100  links. 

The  unit  is  the  chain,  and  the  table  is  made  up  of  divisors  and 
multiples  of  this  unit. 

TABLE. 

7.92  inches  (in.)  make  1  link, 1. 

25  links  "  1  rod, rd. 

4  rods,  or  66  feet,  "  1  chain,  . .  .  ch. 

80  chains  "  1  mile,  ....mi. 


mi. 
1 


ch. 
1 

80 


UNIT    EQUIVALENTS. 

rd. 

4        = 
320      »  = 


i. 
1 

25 
100 

8000 


in. 

7.92 
198 
792 
633GO 


SCALE  — ascending,  7.92,  25,  4,  80;  descending,  80,  4,  25,  7.92. 

NOTE. — The  denomination,  rods,  is  seldom  used  in  chain  measure,  distances 
being  taken  in  chains  and  links. 

II.  SQUARE  MEASURE. 

381.  A  Square  is  a  figure  having  four  equal  sides  and  four 
equal  corners  or  right  angles. 

383.  Area  or  Superficies  is  the  space  or  surface  included 
within  any  given  lines  :  as,  the  area  of  a  square,  of  a  field,  of  a 
board,  etc. 

1  square  yard  is  a  figure  having  four 
sides  of  1  yard  or  3  feet  each,  as  shown 
in  the  diagram.  Its  contents  are  3x3 
=  9  square  feet.  Hence, 

The  contents  or  area  of  a  square,  or 
of  any  other  figure  hatiiKj  d  •uniform 
length  and  a  uniform  breadth,  is  found 
l)ij  multiplying  the  length  ly  the  breadth. 


1  yd.  =  3  ft. 


1  yd.  =  3  ft. 


MEASURES  OF  EXTENSION.  1(37 

Thus,  a  square  foot  is  12  inches  long  and  12  inches  wide,  and  the 
contents  are  12  X  12  —  144  square  inches.  A  board  20  feet  long  and 
10  feet  wide,  is  a  rectangle,  containing  20  x  10  =  200  square  feet. 

The  measurements  for  computing  area  or  surface  are  always 
taken  in  the  denominations  of  linear  measure, 

383.  Square  Measure  is  used  in  computing  areas  or  sur- 
faces ;  as  of  land,  boards,  painting,  plastering,  paving,  etc. 

The  unit  is  the  area  of  a  square  whose  side  is  the  unit  of 
length.  Thus,  the  un,t  of  square  feet  is  1  foot  square;  of  square 
yards,  1  yard  square,  etc. 

TABLE. 

144     square  inches  (sq.  in.)  make  1  square  foot,. . .  .sq.  ft. 

9          "      feet  "      1       "       yard,... sq.  yd. 

30£        "      yards  "      1       "       rod, sq.  rd. 

100          "      rods  "       1  acre, A. 

610     acres  "       1  square  mile, . .  .  sq.  mi. 

UNIT   EQUIVALENTS. 

eq.  ft.  sq.  in. 

eq.yd.                               1  =  144 

eq.rd.                      1     =                  9  =  1296 

A               L    1  =             301  =             2721  =  39204 

sq  mi.          1  =         160  =         4840     =         43560  =  6272640 

1   =  640  =  102400  =  3097GOO     =  27878400  =  4014489600 

SCALE -ascending,  144,  9,  30£,  160,  640;  descending,  640,  160,  301 
9,  144. 

Artificers  estimate  their  work  as  follows : 

By  the  square  foot:  glazing  and  stone-cutting. 

By  the  square  yard :  painting,  plastering,  paving,  ceiling,  and 
paper-hanging. 

By  the  square  of  100  square  feet :  flooring,  partitioning,  roofing, 
slating,  and  tiling. 

Bricklaying  is  estimated  by  the  thousand  bricks,  by  the  square 
yard,  and  by  the  square  of  100  square  feet. 

NOTES. — 1.  In  estimating  the  painting  of  moldings  cornices,  etc.,  the  mea- 
EuriiiLT-line  is  carried  into  all  the  moldings  and  cornices. 

2.  In  estimating  brick-laying  by  either  the   square  yard  or  the  square  of  100 
feet,  the  work  is  understood  to  be  12  inches  or  H  bricks  thick. 

3.  A  thousand  shingl&s  are  estimated  to  cover  1  square,  being  laid  5  inches  tf 
the  weather. 


168  COMPOUND  NUMBERS. 

SURVEYORS'  SQUARE  MEASURE. 

S84.    This  measure  is  used  by  surveyors  in. computing  the 
area  or  contents  of  land. 

TABLE. 

625  square  links  (sq.  1.)  make  1  pole, P. 

16  poles  "      1  square  chain, .  sq.  ch. 

10  square  chains  "      1  acre, A. 

640  acres  "      1  square  mile,  .sq.  mi. 

36  square  miles  (6  miles  square)       "      1  township Tp. 

UNIT    EQUIVALENTS. 


P. 

sq.  ch. 

sq.  1. 

=       625 

A. 

1 

=      16 

=      10000 

sq.  mi. 

1 

=     10 

=     160 

=     100000 

Tp. 

1   = 

640 

=    6400 

=   102400 

=    64000000 

1   = 

36  = 

23040 

==  230400 

=  3686400 

=  2304000000 

SCALE  —  ascending,  625,  16,  10,  640,  36 ;  descending,  36,  640,  10, 
16,  625. 

NOTES. — 1.   A  square  mile  of  land  is  also  called  a  section. 

2.  Canal  and  railroad  engineers  commonly  use  an  engineers'  chain,  which  con- 
sists of  100  links,  each  1  foot  long. 

3.  The  contents  of  land  are  commonly  estimated  in  square  miles,  acres,  and 
hundredths;  the  denomination,  rood,  is  rapidly  going  into  disuse. 

III.  CUBIC  MEASURE. 

28  «5.    A  Cube  is  a  solid,  or  body,  having  six  equal  square 
sides  or  faces. 

28G.    Solidity  is  the  matter  or  space   contained  within   the 
bounding  surfaces  of  a  solid. 

The  measurements  for  computing  solidity  are  always  taken  in 
the  denominations  of  linear  measure. 

If  each  side  of  a  cube  be  1  yard,  or  3 
feet,  1  foot  in  thickness  of  this  cube  will 
contain  3x3x1  =  9  cubic  feet;  and  the 
whole  cube  will  contain  3  x  3  X  3  =  27 
cubic  feet. 

A  solid,   or  body,   may  have  the  three 
dimensions   all   alike   or  all  different.     A 
body  4  ft.  long,  3  ft.  wide,  and  2  ft.  thick 
it.  =  i  yd.  contains  4  X  3  x  2  =  24   cubic  or   solid 

feet.     Hence  we  see  that 


MEASURES  OF  EXTENSION. 

TJie  cubic  or  solid  contents  of  a  body  are  found  ly  multiplying 
the  length,  breadth,  and  thickness  together. 

287.  Cubic  Measure,  also  called  Solid  Measure,  is  used  in 
computing  the  contents  of  solids,  or  bodies;  as  timber,  wood 
stone,  etc. 

The  unit  is  the  solidity  of  a  cube  whose  side  is  the  unit  of 
length.  Thus,  the  unit  of  cubic  feet  is  a  cube  which  measures  1 
foot  on  each  side  ;  the  unit  of  cubic  yards  is  1  cubic  yard,  etc. 

TABLE. 

1728  cubic  inches  (cu.  in.)                   make  1  cubic  foot  .......  cu.  ft. 

27  cubic  feet                                         "       1  cubic  yard  .....  cu.  yd. 

40  cubic  feet  of  round  timber,  or  }     ,                                                m 

50  «           «     hewn        ••           j            1  ton  or  load  .........  T. 

16    cubic  feet  "       1  cord  foot  .......  cd.  ft. 


.....  ca. 

24|  cubic  feet  "      1  Pel, 


SCALE  —  ascending,  1728,  27.  The  other  numbers  are  not  in  a 
regular  scale,  but  are  merely  so  many  times  1  foot.  The  unit  equiva- 
lents, being  fractional,  are  consequently  omitted. 

NOTKS.  —  1.  A  cubic  yard  of  earth  is  called  a  load. 

2.  Railroad  and  transportation  companies  estimate  light  freight  by  the  space 
it  occupies  in  cubic  feet,  and  heavy  freight  by  weight. 

3.  A  pile  of  wood  8  feet  long,  4  feet  wide,  and  4  feet  high,  contains  1  cord; 
and  a  cord  foot  is  1  foot  in  length  of  such  a  pile. 

4.  A  perch  of  stone  or  of  masonry  is  16J  feet  long,  1J  feet  wide,  and  1  foot 
high. 

5.  Joiners,  bricklayers,  and  masons,  make  an  allowance  for  windows,  doors, 
etc.,  of  one  half  the  openings  or  vacant  spaces.     Bricklayers  and  masons,  in  es- 
timating their  work  by  cubic  measure,  make  no  allowance  for  the  corners  of  the 
walls  of  houses,  cellars,  etc.,  but  estimate  their  work  by  the  girt,  that  is,  the 
entire  length  of  the  wall  on  the  outside. 

6.  Engineers,  in  making  estimates  for  excavations  and  embankments,  take  the 
dimensions  with  a  line  or  measure  divided  into  feet  and  decimals  of  a  foot.    The 
computations  are  made  in  feet  and  decimals,  and  the  results  are  reduced  to  cubic 
yards.     In  civil  engineering,  the  cubic  yard  is  the  unit  to  which  estimates  for 
excavations  and  embankments  are  finally  reduced. 

7.  In  scaling  or  measuring  timber  for  shipping  or  freighting,  ^  of  the  solid 
contents  of  round  timber  is  deducted  for  waste  in  hewing  or  sawing.     Thus,  a 
log  that  will  make  40  feet  of  hewn  or  sawed  timber,  actually  contains  50  cubic 
feet  by  measurement;  but  its  market  value  is  only  equal  to  40  cubic  feet  of 
hewn  or  sawed  timber.     Hence,  the  cubic  contents  of  40  feet  of  round  and  50 
feet  of  hewn  timber,  as  estimated  for  market,  are  identical. 

15 


170  COMPOUND  NUMBEKS. 


MEASURES  OF  CAPACITY. 

Capacity  signifies  extent  of  room  or  space. 

389.  Measures  of  capacity  are  all  cubic  measures,  solidity  and 
capacity  being  referred  to  different  units,  as  will  be  seen  by  com- 
paring the  tables. 

Measures  of  capacity  may  be  properly  subdivided  into  two 
classes,  Measures  of  Liquids  and  Measures  of  Dry  Substances. 

I.     LIQUID  MEASURE. 

29O.  Liquid  Measure,  also  called  Wine  Measure,  is  used  in 
measuring  liquids ;  as  liquors,  molasses,  water,  etc. 

The  unit  is  the  gallon,  and  the  table  is  made  up  of  its  divisors 
and  multiples. 

TABLE. 

4    gtlls  (gi.)  make  1  pint, pt. 

2    pints  "  1  quart, qt. 

4    quarts  "  1  gallon, gal. 

31£  gallons  "  1  barrel, bbl. 

2    barrels,  or  63  gal.     "  1  hogshead, ,.  hhd. 


UNIT 

EQUIVALENTS. 

pt. 
qt.                   1 

Ri 

=         4 

gal. 

1 

=        2 

Q 

tibl. 

1 

=       4 

o 

=        32 

1     = 

31£ 

=    126 

=    252 

=    1008 

2    = 

63 

=    252 

=    504 

=    2016 

hhd. 
1 

SCALE  —  ascending,  4,  2,  4,  31£,  2;  descending,  2,  31  £,  4,  2,  4. 

The  following  denominations  are  also  in  use : 

42  gallons  make  1  tierce. 

2  hogsheads,  or  126  gallons,       "       1  pipe  or  butt. 
2  pipes  or  4  hogsheads,  "       1  tun. 

NOTES. — 1.  The  denominations,  barrel  and  hogshead,  are  used  in  estimating 
the  capacity  of  cisterns,  reservoirs,  vats,  etc.  In  Massachusetts  the  barrel  is 
estimated  at  32  gallons. 

2.  The  tierce,  hogshead,  pipe,  butt,  and  tun  are  the  names  of  casks,  and  do 
not  express  any  fixed  or  definite  measures.  They  are  usually  gauged,  and  have 
their  capacities  in  gallons  marked  on  them.  Several  of  these  denominations  aro 
etill  in  use  in  England,  (327 — 330). 


WEIGHTS 
BEER    MEASURE. 

£91.   Beer  Measure  is  a  species  of  liquid  measure  used  in 
measuring  beer,  ale,  and  milk. 


The  unit  is  the  gallon. 


TABLE. 


2    pints  (pt.)  make  1  quart, qt. 

4    quarts  "  1  gallon, gal. 

36    gallons  "  1  barrel, bbl. 

1£  barrels,  or  54  gallons,      "  1  hogshead, .  .hhd. 

UNIT    EQUIVALENTS. 

qt.  pt. 

gal.  1      =  2 

bbi.  1    =        4    =        8 

hhd.          1      =    36    =    144    =    288 

1     =     ij     =    54    =    216    =    432 

SCALE  —  ascending,  2,  4,  36,  H ;  descending,  1£,  36,  4,  2. 

This  measure  is  not  a  standard ;  it  is  rapidly  falling  into  disuse. 

II.     DRY  MEASURE. 

292.  Dry  Measure  is  used  in  measuring  articles  not  liquid ; 
as  grain,  fruit,  salt,  roots,  ashes,  etc. 

The  unit  is  the  bushel,  of  which  all  the  other  denominations  in 
the  table  are  divisors. 

TABLE. 

2  pints  (pt.)       make  1  quart, qt. 

8  quarts  "       1  peck, pk. 

4  pecks  "       1  bushel, . .  bu.  or  bush. 

UNIT   EQUIVALENTS. 

qt.  pt. 

P*.                    1     .   =  2 

bu.                  1        =          8        =  16 

1        =        4        =        32        =  64 
SCALE  —  ascending,  2,  8,  4 ;  descending,  4,  8,  2. 

WEIGHTS. 


T»  JJ  1  \JT  IT. -I  O. 

393.  Weight  is  the  measure  of  the  quantity  of  matter  a  body 
contains,  determined  by  the  force  of  gravity. 

NOTE.  —  The  process  by  which  the  quantity  of  matter  or  the  force  of  gravity 
is  obtained  is  called  Weighing;  and  consists  in  comparing  the  thing  to  be 
weighed  with  some  conventional  standard. 


172  COMPOUND  NUMBERS. 

Three  scales  of  weight  are  used  in  the  United  States  ;  namely, 
Troy,  Avoirdupois,  and  Apothecaries'. 

I.     TROY  WEIGHT. 

394.  Troy  Weight  is  used  in  weighing  gold,  silver,  and 
jewels;  in  philosophical  experiments,  and  generally  where  great 
accuracy  is  required. 

The  unit  is  the  pound,  and  of  this  all  the  other  denominations 
in  the  table  are  divisors. 

TABLE. 

24  grains  (gr.)     make  1  penny  weight,  ..  pwt.  or  dwt. 
20  pennyweights       "     1  ounce,  ----  .............  oz. 

12  ounces  "     1  pound,  ................  Ib. 


UNIT    EQUIVALENTS. 
pwt. 

lb.  1    =      20    =      480 

1    =    12    -=    240    ==    5760 


gr. 

=       24 


SCALE—  ascending,  24,  20,  12;  descending,  12,  20,  24. 
NOTE.  —  Troy  weight  is  sometimes  called  Goldsmiths''  Weight. 

II.     AVOIRDUPOIS  WEIGHT. 
295.   Avoirdupois  Weight  is  used  for  all  the  ordinary  pur- 

poses of  weighing.  • 

The  unit  is  the  pound,  and  the  table  is  made  up  of  its  divisors 
and  multiples. 

TABLE. 

16  ounces  (oz.)    make  1  pound,  ............  lb. 

100  pounds        "   1  hundred  weight,,  .cwt. 
20  cwt.,  or  2000  Ibs.,  "   1  ton,  ...............  T. 

UNIT   EQUIVALENTS. 

lb.  OZ. 

cwt                      1  '  as  .16 

T                1     =       100  =  1600 

1     ==     20     =     2000  =  32000 

SCALE—  ascending,  16,  100,  20  ;  descending,  20,  100,  16. 


WEIGHTS. 


173 


NOTE.  —  The  long  or  gross  ton,  hundred  weight,  and  quarter  were  formerly  in 
common  use;  but  they  are  now  seldom  used  except  in  estimating  English  goods 
at  the  U.  S.  custom-houses,  in  freighting  and  wholesaling  coal  from  the"  Penn- 
sylvania mines,  and  in  the  wholesale  iron  and  plaster  trade. 

LONG  TON  TABLE. 

28  Ib.                           make  1  quarter,                 marked  qr. 

4  qr.    =     112  Ib.       "      1  hundred  -weight,       "  ewt. 

20  cwt.  =  2240  Ib.      "      1  ton,                            "  T. 
SCALE  —  ascending,  28,  4,  20;  descending,  20,  4,  28. 

29G.  The  weight  of  the  bushel  of  certain  grains  and  roots 
has  been  fixed  by  statute  in  many  of  the  States ;  and  these  statute 
weights  must  govern  in  buying  and  selling,  unless  specific  agree- 
ments to  the  contrary  be  made. 


TABLE   OF   AVOIRDUPOIS   POUNDS   IN    A   BUSHEL, 
As  prescribed  by  statute  in  the  several  States  named. 


COMMODITIES. 

.5 

^ 

~n 
0 

Connecticut. 

§ 

ij 
a 

I 

~ 

.!: 
^ 

1 

| 

^ 

M 

S 
« 

12 
3 

to 

• 

I 

A 

& 
£ 

e 

. 

i 

1 

r« 

f 

a 

•s 

1 

5 

5 

>H 

'- 

{ 

1 

« 

c 

2 

• 
> 

| 

1 

z 

8 

2    S 

* 

2 

£ 

K 

c 

50 
40 

52 
32 
54 

CO 

45 

5G 

28 

60 
56 

56 

56 
60 

48 

bo 

14 
40 
46 
60 
•24 
33 
56 
8 
44 
52 
70 
4S 
80 
32 
5T 

60 
54 

45 
60 

2U 

48 
CO 

» 

46 

60 
25 
33 
56 

44 

11 

50 
70 
32 

48 

60 
56 

50 
45 
CO 

4S 
60 
14 

5.: 

46 

% 

: 
s 

68 

35 

57 

60 
56 

50 
45 
CO 
20 

48 
60 
14 
52 

60 
56 

« 

56 
50 

33^ 
57 

60 
56 

50 
45 
60 
20 

32 

56 
32 
32 

60 

11 

50 
30 

CO 
50 

4C 
46 

„ 

50 

30 
52 

56 
50 

CO 

48 

42 

CO 

28 

2S 

56 
32 
56 

60 

48 

42 

CO 
28 
28 

5C 
32 

56 
CO 

48 
CO 
14 
52 
46 
60 
24 
33 
56 

44 

52 

feO 
35 
57 

CO 
5C 

50 

45 
CO 
20 

' 

30 
60 

48 

50 
64 

55 

56 

30 

60 
56 

60 

48 
62 

48 
00 

05 
58 

32 

60 
60 
56 

56 
44 
60 

48 

60 
56 

56 

32 
56 
CO 

46 

42 

CO 

28 

28 

56 

34 

00 

36 

60 

47 
48 

56 
32 

56 
60 

50 

50 
CO 
M 

46 
46 

56 

32 

60 
56 

60 

45 

42 

60 
28 

28 

56 

36 
50 

60 
56 

60 

Blue  Grass  Seed 
Buckwheat  
Castor  Beans  
Clover  Seed  
Dried  Apples  
Dried  Peaches- 
Flax  Seed  
Hair 

Hemp  Seed  
Indian  Corn  
Ind.  Corn  in  ear 
Ind.Corn  Meal. 
Mineral  Coal1... 
Oats  

Onions  

Peas  

Potatoes  

Rye  

Eye  Meal  
Salt*  

Timothy  Seed... 
Wheat  
Wheat  Bran  

1  In  Kentucky,  80  Ibs.  of  bituminous  coal  or  70  Ibs.  of  cannel  coal  make  1  bushel. 
a  In  Pennsylvania,  80  Ibs.  coarse,  70  Ibs.  ground,  or  62  Ibs.  fine  salt  make  1  bushel ;  and 
in  Illinois,  50  Ibs.  common  or  55  Ibs.  fine  salt  make  1  bushel. 
*  In  Maine,  64  Ibs.  of  ruta  baga  turnips  or  beets  make  1  bushel. 

15  * 


•[74  COMPOUND  NUMBERS. 

NOTES.  —  1.  The  weight  of  a  barrel  of  flour  is  7  quarters  of  old,  or  long  ton 
Weight. 

2.  The  weight  of  a  bushel  of  Indian  corn  and  rye,  as  adopted  by  most  of  the 
States,  and  of  a  bushel  of  salt  is  2  quarters  ;  and  of  a  barrel  of  salt  10  quarters, 
or  i  of  a  long  ton. 

The  following  denominations  are  also  in  use : 
56  pounds  make  1  firkin  of  butter. 


100 
100 
196 
200 

280 


1  quintal  of  dried  salt  fish. 

1  cask  of  raisins. 

1  barrel  of  flour. 

1       "      "  beef,  pork,  or  fish. 

1       "      "  salt  at  the  N.  Y.  State  salt  works. 


III.  APOTHECARIES'  WEIGHT. 

Apothecaries'  Weight  is  used  by  apothecaries  and  phy- 
sicians in  compounding  medicines ;  but  medicines  are  bought  and 
sold  by  avoirdupois  weight. 

The  unit  is  the  pound,  of  which  all  the  other  denominations  in 
the  table  are  divisors. 

TABLE. 

20  grains  (gr.)  make  1  scruple, so.  or  9. 

3  scruples  "     1  dram, dr.  or  3. 

8  drams  "     1  ounce, oz.  or  ^. 

12  ounces  "     1  pound, Ib.  or  ft). 

UNIT    EQUIVALENTS. 

SC.  gr. 

dr.  1     =  20 

.     oz.  1    =       3    =       60 

lb.  1    =      8    =      24    =     480 

1    =    12    =    96    =    288    =    5760 
SCALE  —  ascending,  20,  3,  8,  12;  descending,  12,  8,  3,  20. 

APOTHECARIES'  FLUID  MEASURE. 

£O8.  The  measures  for  fluids,  as  adopted  by  apothecaries  and 
physicians  in  the  United  States,  to  be  used  in  compounding  medi- 
cines, and  putting  them  up  for  market,  are  given  in  the  following 

TABLE. 

60  minims,  (^l)  make  1  fluidrachm, f%. 

8  fluidrachms,  "  1  fluidounce, f.^. 

16  fluidounces,  "  1  pint, 0. 

8  pints,  "  1  gallon, Cong. 


TIME. 


175 


UNIT   EQUIVALENTS. 
f5                         1H 

f3 

1    = 

60 

0. 

1 

=          8    = 

480 

Cong. 

1     = 

16 

=      128    = 

7680 

i      

8    = 

128 

=    2048    = 

61440 

SCALE  —  ascending,  60,  8,  16,  8 ;  descending,  8,  16,  8,  60. 

MEASURE  OF  TIME. 

299.   Time  is  the  measure  of  duration.     The  unit  is  the  day, 
and  the  table  is  made  up  of  its  divisors  and  multiples. 


TABLE. 


60  seconds  (sec.)  make 
60  minutes, 
24  hours, 
7  days, 

365  days, 

366  days, 

12  calendar  months, 
100  years, 


1  minute, min. 

1  hour, h. 

1  day, da. 

1  week, wk. 

1  common  year, yr. 

1  leap  year, yr. 

1  year, yr. 

1  century, C. 


wk. 
1 


UNIT   EQUIVALENTS. 

min. 

h.  1 

da.  1     =  60 

1  =  24  =  1440 
7  =  168  =  10080 
8760  =  525600 


sec. 

=  60 

=  3600 

=  86400 

=  604800 

=  31536000 

=  31622400 


yr.  mo.  f  365     = 

1    =    12  :    (366    =    8784    =    527040 

SCALE  —  ascending,  60,  60,  24,  7 ;  descending,  7,  24,  60,  60. 

The  calendar  year  is  divided  as  follows : — 


No.  of  month. 

1 
9 

3 

4 

5 

6 

7 

8 

9 

10 
11 
12 


Season. 

Winter, 
Spring, 


Summer, 


Autumn, 
Winter, 


Names  of  months. 

Abbreviations- 

No  of  days. 

January, 

Jan. 

31 

February, 

Feb. 

28  or  29 

March, 

Mar. 

31 

April, 
May, 

Apr. 

30 
31 

June, 

Jun. 

30 

July, 



31 

August, 

Aug. 

31 

September, 

Sept. 

30 

October, 

Oct. 

31 

November, 

Nov. 

30 

December, 

Dec. 

31 

NOTES.  —  1.  In  most  business  transactions  30  days  are  called  1  month. 

2.  The  civil  day  begins  and  ends  at  12  o'clock,  midnight.     The  astrouomi- 


176  COMPOUND  NUMBERS. 

i 

cal  day,  used  by  astronomers  in  dating  events,  begins  and  ends  at  12  o'clock, 
noon.     The  civil  year  is  composed  of  civil  days. 


BISSEXTILE    OR    LEAP   YEAR. 

3OO.  The  period  of  time  required  by  the  sun  to  pass  from 
one  vernal  equinox  to  another,  called  the  vernal  or  tropical  year, 
is  exactly  365  da.  5  h.  48  min.  49.7  sec.  This  is  the  true  year, 
and  it  exceeds  the  common  year  by  5  h.  48  min.  49.7  sec. 

If  365  days  be  reckoned  as  1  year,  the  time  lost  in  the  calendar 
will  be 

In  1  yr.,        5  h.  48  min.  49.7  sec. 

"  4  *"         23  "  15     "     18.8   " 

The  time  thus  lost  in  4  years  will  lack  only  44  min.  41.2  sec.  of 
1  entire  day.  Hence, 

If  every  fourth  year  be  reckoned  as  leap  year,  the  time  gained  in 
the  calendar  will  be, 

In      4  yr.,  44  min.  41.2  sec. 

"100"    (  =  25X4    yr.)     18  h.  37     "     10      " 

The  time  thus  gained  in  100  years  will  lack  only  5  h.  22  min.  50 
sec.  of  1  day.  Hence 

If  every  fourth  year  be  reckoned  as  leap  year,  the  centennial  years 
excepted,  the  time  lost  in  the  calendar  will  be, 

In  100  yr.,        5  h.  22  min.  50  sec. 
"  400  "         21  "  31     "     20   " 

The  time  thus  lost  in  400  years  lacks  only  2  h.  28  min.  40  sec.  of  1 
day.  Hence 

If  every  fourth  year  be  reckoned  as  leap  year,  3  of  every  4  cen- 
tennial years  excepted,  the  time  gained  in  the  calendar  will  be, 

In    400  yr.,        2  h.  28  min.  40  sec. 
"  4000  "         24  h.  46  min.  40  sec. 

The  following  rule  for  leap  year  will  therefore  render  the  calendar 
correct  to  within  1  day,  for  a  period  of  4000  years. 

I.  Every  year  that  is  exactly  divisible  by  4  is  a  leap  year, 
the  centennial  years  excepted ;  the  other  years  are  common  years. 

II.  Every  centennial  year  that  is  exactly  divisible  by  400  is  a 
leap  year ;  the  other  centennial  years  are  common  years. 

NOTES.  —  1.  Julius  Caesar,  the  Roman  Emperor,  decreed  that  the  year  should 
consist  of  365  days  6  hours;  that  the  fi  hours  should  be  disregarded  for  3  suc- 
cessive years,  and  an  entire  day  be  added  to  every  fourth  year.  This  day  was 
inserted  in  the  calendar  between  the  24th  and  25th  days  of  February,  and  is 
called  the  intercalary  day.  As  the  Romans  counted  the  days  backward  from  the 
first  day  of  the  following  month,  the  24th  of  February  was  called  by  them  sexto 


CIRCULAR  MEASURE.  177 

calendas  Martii,  the  sixth  before  the  calends  of  March.  The  intercalary  day 
which  followed  this  was  called  bit-sexto  calendas  Martii;  hence  the  name 
bissextile. 

2.  In  1582  the  error  in  the  calendar  as  established  by  Julius  Caesar  had  in- 
creased to  10  days;  that  is,  too  much  time  had  been  reckoned  as  a  year,  until 
the  civil  year  was  10  days  behind  the  solar  year.     To  correct  this  error,  Pope 
Gregory  decreed  that  10  entire  days  should  be  stricken  from  the  calendar,  and 
that  tlTe  day  following  the  3d  day  of  October,  1582,  should  be  the  14th.     Thi« 
brought  the  vernal  equinox  at  March  21  —  the  date  on  which  it  occurred  in  the 
year  325,  at  the  time  of  the  Council  of  Nice. 

3.  The  year  as  established  by  Julius  Caesar  is  sometimes  called  the  Julian 
year  ;  and  the  period  of  time  in  which  it  was  in  force,  namely  from  46  years 
B.  C.  to  1582,  is  called  the  Julian  Period. 

4.  The  year  as  established  by  Pope  Gregory  is  called  the  Gregorian  year,  and 
the  calendar  now  used  is  the  Gregorian  Calendar. 

5.  Most  Catholic  countries  adopted  the  Gregorian  Calendar  soon  after  it  was 
established.    Great  Britain,  however,  continued  to  use  the  Julian  Calendar  until 
1752.     At  this  time  the  civil  year  was  11  days  behind  the  solar  year.     To  cor- 
rect this  error,  the  British  Government  decreed  that  11  days  should  be  stricken 
from  the  calendar,  and  thrt  the  day  following  the  2d  day  of  September,  1752, 
should  be  the  14th. 

6.  Time  before  the  adoption  of  the   Gregorian  Calendar  is  called  Old  Style 
(0.  S),  and  since,  New  Style,  (N.  S.)     In  Old  Style  the  year  commenced  March 
25,  and  in  New  Style  it  commences  January  1. 

7.  Russia  still  reckons  time  by  Old  Style,  or  the  Julian  Calendar;  hence  their 
dates  are  now  12  days  behind  ours. 

8.  The  centuries  are  numbered  from  the  commencement  of  the  Christian  era; 
the  months  from  the  commencement  of  the  year;  the  days  from  the  commence- 
ment of  the  month,  and  the  hours  from   the  commencement  of  the  day,  (12 
o'clock,  midnight.)     Thus,  May  23,  1860,  9  o'clock  A.M.,  is  the  9th  hour  of  the 
23d  day  of  the  5th  month  of  the  60th  year  of  the  19th  century. 

MEASURE   OF   ANGLES. 

3O1.  Circular  Measure,  or  Circular  Motion,  is  used  princi- 
pally in  surveying,  navigation,  astronomy,  and  geography,  for 
reckoning  latitude  and  longitude,  determining  locations  of  places 
and  vessels,  and  computing  difference  of  time. 

Every  circle,  great  or  small,  is  divisible  into  the  same  number 
of  equal  parts :  as  quarters,  called  quadrants;  twelfths,  called  signs; 
360ths,  called  degrees,  etc.  Consequently  the  parts  of  different 
circles,  although  having  the  same  names,  are  of  different  lengths. 

The  unit  is  the  degree,  which  is  -g^  part  of  the  space  about  a 
point  in  any  plane.  The  table  is  made  up  of  divisors  and  multiples 
of  this  unit. 

TABLE. 

60  seconds  ("}       make  1  minute, . .  . . '. 
60  minutes  "      1  degree,....0. 

30  degrees  "      1  sign, S. 

12  signs,  or  360°,       "      i  circle, C. 

M 


178  COMPOUND  NUMBERS. 

( 

UNIT    EQUIVALENTS. 

I  // 

1  =  60 

s             l   =        60  =  3600 

c            l   =     30   =     1800  =  108000 

1   =   12  =  360   =  21600  =  1296000 

SCALE  —  ascending,  60,  60,  30,  12 ;  descending,  12,  30,  60,  60. 

NOTES.  —  1.  Minutes  of  the  earth's  circumference  are  called  geographic  or 
nautical  miles. 

2.  The  denomination,  signs,  is  confined  exclusively  to  Astronomy. 

3.  A  degree  has  no  fixed  linear  extent.    When  applied  to  any  circle  it  is  always 
j^  part  of  the  circumference.     But,  strictly  speaking,  it  is  not  any  part  of  a 
circle. 

4.  90°  make  a  quadrant  or  right-angle; 
60°       «      "  sextant      "   &  of  a  circle. 


MISCELLANEOUS   TABLES. 
3O2.    COUNTING. 

12  units  or  things  make  1  dozen. 
12  dozen  "      1  gross. 

12  gross  "      1  great  gross. 

20  units  "      1  score. 

303.  PAPER. 

24  sheets make 1  quire. 

20  quires  "  1  ream. 

2  reams  "  1  bundle. 

5  bundles  "  1  bale. 

304.  BOOKS. 

The  terms  folio,  quarto,  octavo,  duodecimo,  etc.,  indicate  the 
number  of  leaves  into  which  a  sheet  of  paper  is  folded. 

A  sheet  folded  in    2  leaves  is  called  a  folio. 

A  sheet  folded  in    4  leaves  "  a  quarto,  or  4to. 

A  sheet  folded  in     8  leaves  "  an  octavo,  or  8vo. 

A  sheet  folded  in  12  leaves  "  a  12mo. 

A  sheet  folded  in  16  leaves  "  a  16mo. 

A  sheet  folded  in  18  leaves  "  an  18mo. 

A  sheet  folded  in  24  leaves  "  a  24mo. 

A  sheet  folded  in  32  leaves  "  a  32mo. 

3O5.    COPYING. 

72  words  make  1  folio  or  sheet  of  common  law. 
90       "          "      1     "       *'       "       "  chancery. 


GOVERNMENT  STANDARDS.  179 


GOVERNMENT  STANDARDS 
OF    MEASURES     AND   WEIGHTS. 

306.  In  early  times,  almost  every  province  and  chief  city  had 
its  own  measures  and  weights  ;  but  these  were  neither  definite  nor 
uniform.     This  variety  in  the  weights  and  measures  of  different 
countries  has  always  proved  a  serious  embarrassment  to  commerce  ; 
hence  the  many  attempts  that  have  been  made  in  modern  times  to 
establish  uniformity. 

The  English,  American,  and  French  Governments,  in  establish- 
ing their  standards  of  measures  and  weights,  founded  them  upon 
unalterable  principles  or  laws  of  nature,  as  will  be  seen  by  ex- 
amining the  several  standards. 

UNITED  STATES  STANDARDS. 

307.  In  the  year  1834  the  U.  S.  Government  adopted  a  uni- 
form standard  of  weights  and  measures,  for  the  use  of  the  custom 
houses,  and  the  other  branches  of  business  connected  with  the 
General  Government.     Most  of  the  States  which  have  adopted 
any  standards  have  taken  those  of  the  General  Government. 

308.  The  invariable  standard  unit  from  which  the  standard 
units  of  measure  and  weight  are  derived  is  the  day. 

Astronomers  have  proved  that  the  diurnal  revolution  of  the 
earth  is  entirely  uniform,  always  performing  equal  parts  of  a  revo- 
lution on  its  axis  in  equal  periods  of  duration. 

Having  decided  upon  the  invariable  standard  unit,  a  measure 
of  this  unit  was  sought  that  should  in  some  manner  be  connected 
with  extension  as  well  as  with  this  unit.  A  clock  pendulum 
whose  rod  is  of  any  given  length,  is  found  always  to  vibrate  the 
Bame  number  of  times  in  the  same  period  of  duration.  Having 
now  the  day  and  the  pendulum,  the  different  standards  hereafter 
have  been  determined  and  adopted. 


STANDARD    OF   EXTENSION. 

3OO.    The  U.  S.  standard  unit  of  measures  of  extension)  whether 
linear,  superficial,  or  solid,  is  the  yard  of  3  feet,  or  36  inches, 


180  COMPOUND  NUMBERS. 

» 

and  is  the  same  as  the  Imperial  standard  yard  of  Great  Britain. 
It  is  determined  as  follows  :  The  rod  of  a  pendulum  vibrating 
seconds  of  mean  time,  in  the  latitude  of  London,  in  a  vacuum,  at 
the  level  of  the  sea,  is  divided  into  391393  equal  parts,  and  360000 
of  these  parts  are  36  inches,  or  1  standard  yard.  Hence,  such  a 
pendulum  rod  is  39.1393  inches  long,  and  the  standard  yard  is 
ISfiBi  of  the  length  of  the  pendulum  rod. 

STANDARDS    OF  CAPACITY. 

310.  The  U.  S.  standard  unit  of  liquid  measure  is  the  old 
English   wine    gallon,   of  231    cubic  inches,  which  is   equal  to 
8.33888  pounds  avoirdupois  of  distilled  water  at  its  maximum 
density;  that  is,  at  the  temperature  of  39.83°  Fahrenheit,  the  ba- 
rometer at  30  inches. 

311.  The  U.  S.  standard  unit  of  dry  measure  is  the  British 
Winchester  bushel,  which  is  18 £  inches  in  diameter  and  8  inches 
deep,  and  contains  2150.42  cubic  inches,  equal  to  77.6274  pounds 
avoirdupois  of  distilled  water,  at  its  maximum  density.     A  gallon, 
dry  measure,  contains  268.8  cubic  inches. 

NOTES. — 1.  Grain  and  some  other  commodities  are  sold  by  stricken  measure, 
and  in  such  cases  the  "  measure  is  to  be  stricken  with  a  round  stick  or  roller, 
straight,  and  of  the  same  diameter  from  end  to  end." 

2.  Coal,  ashes,  marl,  manure,  corn  in  the  ear,  fruit  and  roots  are  sold  by  heap 
measure.     The  bushel,  heap  measure,  is  the  Winchester  bushel  heaped  in  the 
form  of  a  cone,  which   cone  must  be  19J  inches  in  diameter  (=  to  the  outside 
diameter  of  the  standard  bushel  measure,)  and  at  least  6  inches  high.     A  bushel, 
heap  measure,  contains  2747.7167  cubic  inches,  or  597.2967  cubic  inches  more 
than  a  bushel  stricken  measure.    Since  1  peck  contains  iis^o.4  =  537.605  cubic 
inches,  the  bushel,  heap  measure,  contains  59.6917  cubic  inches  more  than  5 
pecks.     As  this  is  about  1  bu.  1  pk.  If  pt.,  it  is  sufficiently  accurate  in  practice, 
to  call  5  pecks  stricken  measure  a  heap  bushel. 

3.  A  standard  bushel,  stricken   measure,  is  commonly    estimated    at   2150.4 
cubic  inches.     The  old  English  standard  bushel  from  which  the  United  States 
standard  bushel  was  derived,  was  kept  at  Winchester,  England;  hence  the  name. 

4.  The  wine  and  dry  measures  of  the  same  denomination  are  of  different  capac- 
ities.      The  exact  and  the  relative  size  of  each  may  be  readily  seen  by  the  fol' 
lowing 

512.     COMPARATIVE   TABLE   OF   MEASURES    OF   CAPACITY. 

Cubic  in.  in        Cubic  in.  in         Cubic  in.  in        Cubic  in.  in 
one  gallon,          one  quart.  one  pint.  one  gill. 

AVine  measure, 231       57f       28f       7^ 

Dry  measure  (^pk.,)..  268J      67£       33s 

NOTE.    —    The  beer  gallon  of  282  inches  is  retained  in  use  onl.v  by  custom. 


GOVERNMENT  STANDARDS. 


STANDARDS    OF   WEIGHT. 

3  l«t.  It  has  been  found  that  a  given  volume  or  quantity  of 
distilled  rain  water  at  a  given  temperature  always  weighs  the  same. 
Hence,  a  cubic  inch  of  distilled  rain  water  has  been  adopted  as 
the  standard  of  weight. 

314.  The  U.  S.  standard  unit  of  weight  is  the  Troy  pound  of 
the  Mint,  which  is  the  same  as  the  Imperial  standard  pound  of 
Great  Britain,  and  is  determined  as  follows  :  A  cubic  inch  of  dis- 
tilled water  in  a  vacuum,  weighed  by  brass  weights,  also  in  a 
vacuum,  at  a  temperature  of  62°  Fahrenheit's  thermometer,  is 
equal  to  252.458  grains,  of  which  the  standard  Troy  pound  con- 
tains 5760. 

31d.  The  U.  S.  Avoirdupois  pound  is  determined  from  the 
standard  Troy  pound,  and  contains  7000  Troy  grains.  Hence, 
the  Troy  pound  is  f  J§§  =  j^l  of  an  avoirdupois  pound.  But 
the  Troy  ounce  contains  5yf  °  =  480  grains,  and  the  avoirdupois 
ounce  7fig°  =  437.5  grains;  and  an  ounce  Troy  is  480  —  437.5 
=  42.5  grains  greater  than  an  ounce  avoirdupois.  The  pound, 
ounce,  and  grain,  Apothecaries'  weight,  are  the  same  as  the  like 
denominations  in  Troy  weight,  the  only  difference  in  the  two 
tables  being  in  the  divisions  of  the  ounce. 

316.     COMPARATIVE    TABLE    OF    WEIGHTS. 

Troy.  Avoirdupois.  Apothecaries'. 

1  pound     =     5760  grains,     =     7000  grains.    =     5760  grains, 
1  ounce     =      480      "  =      437.5    "        =      480       " 

175  pounds,   =       144  pounds.  =       175  pounds, 

STANDARD    SETS    OF    WEIGHTS    AND    MEASURES. 

317.  A  uniform  set  of  weights  and  measures  for  all  the  States 
was  approved  by  Congress,  June  14,  1836,  and  furnished  to  the 
States  in  1842.  The  set  furnished  consisted  of 

A  yard. 

A  set  of  Troy  weights. 

A  set  of  Avoirdupois  weights. 


132  COMPOUND   NUMBERS. 

A  wine  gallon,  and  its  subdivisions. 
A  half  bushel,  and  its  subdivisions. 

318*    State  Sealers  of  Weights  and  Measures  furnish  standard 
sets  of  weights  and  measures  to  counties  and  towns. 
A  County  Standard  consists  of 

1.  A  large  balance,  comprising  a  brass  beam  and  scale  dishes, 
with  stand  and  lever. 

2.  A  small  balance,  with  a  drawer  stand  for  small  weights. 

3.  A  set  of  large  brass  weights,  namely,  50,  20,  10,  and  5  Ib. 

4.  A  set  of  small  brass  weights,  avoirdupois,  namely,  4,  2,  and 
L  Ib.,  8,  4,  2,  1,  J,  and  i  oz. 

5.  A  brass  yard  measure,  graduated  to  feet  and  inches,  and  the 
first  foot  graduated  to  eighths  of  an  inch,  and  also  decimally ;  with 
a  graduation  to  cloth  measure  on  the  opposite  side ;  in  a  case. 

6.  A  set  of  liquid  measures,  made  of  copper,  namely,  1  gal.,  £ 
gal.,  1  qt.,  1  pt.,  £  pt.,  1  gi.;  in  a  case. 

7.  A  set  of  dry  measures,  of  copper,  namely,  £  bu.,  1  pk.,  £  pk. 
(or  1  gal.),  2  qt.  (or  £  gal.),  1  qt.;  in  a  case. 

ENGLISH  MEASURES  AND  WEIGHTS. 

GOVERNMENT    STANDARDS. 

31Q0  The  English  act  establishing  standard  measures  and 
weights,  called  "  The  Act  of  Uniformity,"  took  effect  Jan.  1, 1826, 
and  the  standards  then  adopted,  form  what  is  called  the  Imperial 
System. 

32O.  The  Invariable  Standard  Unit  of  this  system  is  the 
same  as  that  of  the  United  States,  and  is  described  in  the  Act  of 
Uniformity  as  follows:  "Take  a  pendulum  which  will  vibrate 
seconds  in  London,  on  a  level  of  the  sea,  in  a  vacuum;  divide  all 
that  part  thereof  which  lies  between  the  axis  of  suspension  and 
the  center  of  oscillation,  into  391393  equal  parts ;  then  will  10000 
of  those  parts  be  an  imperial  inch,  12  whereof  make  a  foot,  and 
36  whereof  make  a  yard." 


ENGLISH  MEASURES  AND  WEIGHTS.  183 

STANDARD    OF   EXTENSION. 

The  English  Standard  Unit  of  Measures  of  Extension, 
whether  linear,  superficial,  or  solid,  is  identical  with  that  of  the 
United  States,  (3O9). 

STANDARDS    OF   CAPACITY. 

322.  The  imperial  Standard  Gallon,  for  liquids  and  all  dry 
substances,  is  a  measure  that  will  contain  10  pounds  avoirdupois 
weight  of  distilled  water,  weighed  in  air,  at  62°  Fahrenheit,  the 
barometer  at  30  inches.     It  contains  277.274  cubic  inches. 

323.  The  Imperial  Standard  Bushel  is  equal  to  8  gallons  or 
80  pounds  of  distilled  water,  weighed  in  the  manner  above  de- 
scribed.    It  contains  2218.192  cubic  inches. 

STANDARDS   OF   WEIGHT. 

324.  The  Imperial  Standard  Pound   is    the    pound   Troy, 
which  is  identical  with  that  of  the  United  States  Standard  Troy 
pound  of  the  Mint,  (314.) 

32o.  The  Imperial  Avoirdupois  Pound  contains  7000  Troy 
grains,  and  the  Troy  pound  5760.  It  also  is  identical  with  the 
United  States  avoirdupois  pound. 

TABLES. 

326.  The  denominations  in  the  standard  tables  of  measures 
of  extension,  capacity,  and  weights,  are  the  same  in  Great  Britain 
and  the  United  States.  But  some  denominations  in  several  of  the 
tables  are  in  use  in  various  parts  of  Great  Britain  that  are  not 
known  in  the  United  States. 

These  denominations  are  retained  in  use  by  common  consent, 
and  are  recognized  by  the  English  common  law.  They  are  as  fol- 
lows: 

327.    MEASURES    OF   EXTENSION. 

18  inches  make  1  cubit. 

45  inches  or  1        «     -i    11 

5  quarters  of  the  standard  yard  j 

NOTE. — The  cubit  was  originally  the  length  of  a  man's  forearm  and  hand;  or 
the  distance  from  the  elbow  to  the  e?id  of  the  middle  finger. 


184  COMPOUND  NUMBERS. 

328.     MEASURES    OF    CAPACITY. 


LIQUID    MEASURES. 


9  old  ale  gallons  make  1  firkin. 

4  firkins  "      1  barrel  of  beer. 

£  Imperial    "  "1  firkin. 
Imperial  gallons  or 


63  wine 

70  Imperial  gallons  or 

84  wine  " 

2  hogsheads,  that  is 
105  Imperial  gallons  or 
126  wine 
2  pipes 


1  hogshead. 

1  puncheon  or 
£  of  a  tun. 

1  pipe. 
1  tun. 


Pipes  of  wine  are  of  different  capacities,  as  follows : 

110  wine  gallons  make  1  pipe  of  Madeira. 

f  Barcelona, 
120      "         "  1     "       \  Vidonia,  or 

I  Teneriffe. 

130      "         "  1     "          Sherry. 

138      "         "  1     "          Port, 

14.0      tt         n  i     «        (Bucellas,  or 

|  Lisbon. 

329.     DRY    MEASURE. 

8  bushels  of  70  pounds  each  make  1  quarter  of  wheat. 
36       "        heaped  measure,        "      1  chaldron  of  coal. 

NOTE. — The  quarter  of  wheat  is  560  pounds,  or  J  of  a  ton  of  2240  pounds. 

33O.     WEIGHTS. 

8  pounds  of  butchers'  meat  make  1  stone. 

14        "        "   other  commodities  "      1      "     or  £  of  a  cwt. 

2  stone,  or  28  pounds  "      1  todd  of  wool. 

70  pounds  of  salt  "      1  bushel. 

NOTE. — The  English  quarter  is  28  pounds,  the  hundred  weight  is  112  pounds, 
and  the  ton  is  20  hundred  weight,  or  2240  pounds. 


FRENCH  MEASURES  AND  WEIGHTS. 
GOVERNMENT    STANDARDS. 

331.  The  tables  of  standard  measures  and  weights  adopted 
by  the  French  Government  are  all  formed  upon  a  decimal  scale, 
and  constitute  what  is  called  the  French  Metrical  System. 


FRENCH  MEASURES  AND  WEIGHTS.  185 

332.  Invariable  Standard  Unit.     The  French  metrical  sys- 
tem has,  for  its  unit  of  all  measures,  whether  of  length,  area, 
solidity,  capacity,  or  weight,  a  uniform  invariable  standard,  adopted 
from  nature  and  called  the  mitre.     It  was  determined  and  estab- 
lished as  follows  :  a  very  accurate  survey  of  that  portion  of  the 
terrestrial  meridian,  or  north  and  south  circle,  between  Dunkirk 
and  Barcelona,  France,  was  made,  under  the  direction  of  Govern- 
ment, and  from  this  measurement  the  exact  length  of  a  quadrant 
of  the  entire  meridian,  or  the  distance  from  the  equator  to  the 
north  pole,  was  computed.    The  ten  millionth  part  of  this  arc  was 
denominated  a  metre,  and  from  this  all  the  standard  units  of 
measure  and  weight  are  derived  and  determined. 

STANDARDS    OF    EXTENSION. 

333.  ^e  French   Standard  Linear  Unit  is  the  me*tre. 

334.  The  French   Standard  Unit  of  Area  is  the  Are,  which 
is  a  unit  10  metres  square,  and  contains  100  square  metres. 

335.  The  French  Standard  Unit  of  Solidity  and   Capacity 
is  uhe  Litre,  which  is  the  cube  of  the  tenth  part  of  the  me'tre. 

STANDARD    OF   WEIGHT. 

33G  The  French  Standard  Unit  of  Weight  is  the  Gramme* 
which  is  determined  as  follows :  the  weight  in  a  vacuum  of  a 
cubic  decimetre  or  litre  of  distilled  water,  at  its  maximum  density, 
was  called  a  kilogramme,  and  the  thousandth  part  of  this  was 
called  a  Gramme,  and  was  declared  to  be  the  unit  of  weight. 

NOMENCLATURE    OF    THE    TABLES. 

337.  It  has  already  been  remarked,  (331 ),  that  the  tables  are 
all  formed  upon  a  decimal  scale.  The  names  of  the  multiples  and 
divisors  of  the  Government  standard  units  in  the  tables  are  formed, 
by  combining  the  names  of  the  standard  units  with  prefixes ;  the 
names  of  the  multiples  being  formed  by  employing  the  prefixes 
deca,  (ten),  hecto,  (hundred),  kilo,  (thousand),  and  myria,  (ten 
thousand),  taken  from  the  Greek  numerals ;  and  the  names  of  the 
divisors  by  employing  the  prefixes  deci,  (tenth),  centi,  (hundredth), 
16* 


186  COMPOUND  NUMBERS. 

mili,  (thousandth),  from  the  Latin  numerals.  Hence  the  name 
of  any  denomination  indicates  whether  a  unit  of  that  denomination 
is  greater  or  less  than  the  standard  unit  of  the  table. 

338.  I.  FRENCH  LINEAR  MEASURE. 

TABLE. 


10  millimetres   ma 
10  centimetres       ' 
10  decimetres        ' 
10  metres               ' 
10  decametres       * 
10  hectometres       ' 
10  kilometres        ' 

ke  1  centimetre. 
1  decimetre. 
1  metre. 
1  decametre. 
1  hectometre. 
1  kilometre. 
1  myriametre. 

NOTES.— 1.  The  metre  is  equal  to  39.3685  inches,  the  standard  rod  of  brasi 
on  which  the  former  is  measured  being  at  the  temperature  of  32°  Fahrenheit, 
and  the  English  standard  brass  yard  or  "  Scale  of  Troughton"  at  62°.  Hence,  a 
metre  is  equal  to  3.2807  feet  English  measure. 

2.  The  length  of  a  metre  being  39.3685  inches,  and  of  a  clock  pendulum 
vibrating  seconds  at  the  level  of  the  sea  in  the  latitude  of  London  39.1393 
inches,  the  two  standards  differ  only  .2292,  or  less  than  £  of  an  inch. 

339.  II.  FRENCH  SQUARE  MEASURE. 

TABLE. 

100  square  metres,  or  centiares  (10  metres  square)  make  1  are. 
100  ares  (10  ares  square)  "      1  hectoare. 

NOTE.  —  A  square  metre  or  centiare  is  equal  to  1.19589444  square  yards,  and 
an  are  to  119.589444  square  yards. 

34O.  III.  FRENCH  LIQUID  AND  DRY  MEASURE. 

TABLE. 

10  decilitres  make  1  litre. 

10  litres  "  1  decalitre. 

10  decalitres       "  1  hectolitre. 

10  hectolitres      "  1  kilolitre. 

NOTES.— 1.  A  litre  is  equal  to  61.53294  cubic  inches,  or  1.06552  quarts  of  a  U. , 
S.  liquid  gallon. 

2.  A  table  of  Solid  or  Cubic  Measure  is  also  in  use  in  some  parts  of  France, 
although  it  is  not  established  or  regulated  by  government  enactments  or  decrees. 
The  unit  of  this  table  is  a  cubic  metre,  which  is  equal  to  61532.94238  cubic 
inches,  or  35.60934  cubic  feet.  This  unit  is  called  a  Stere. 

TABLE. 

10  decisteres  make  1  stere. 
10  steres  "       I  decastere. 


MONEY  AND  CURRENCIES.  187 

341.  IV.  FRENCH  WEIGHT. 


T 

10  milligrammes   nn 
10  centigrammes 
10  decigrammes 
10  grammes 
10  decagrammes 
10  hectogrammes 
100  kilogrammes 

10  quintals                ' 

ABLE. 

ike      1  centigramme. 
1  decigramme. 
1  gramme. 
1  decagramme. 
1  hectogramme. 
1  kilogramme. 
1  quintal. 
«        (  1  millier,  or 
{  1  ton  of  sea  wa 

NOTES. — 1.  A  gramme  is  equal  to  15.433159  Troy  grains. 

2.   A  kilogramme  is  equal  to  2  Ib.  8  oz.  3  pwt.  1.159  gr.  Troy,  or  2  Ib.  3  oz. 
4.1549  dr.  Avoirdupois. 

342.   COMPARATIVE  TABLE  OF  THE  UNITED  STATES,  ENGLISH, 
AND  FRENCH  STANDARD  UNITS  OF  MEASURES  A«ND  WEIGHTS. 

United  States.  English.  French. 

Extension,    Yd.  of  3  ft.,  or  36  in.  Same  as  U.  S.  Metre,  39.3685  in. 

n        .,        )  Wine  gal.,  231  cu.  in.  Imp'l  gal.,  277.274 cu.  in.    Litre, '61.53294  cu.  in. 

*ciiy,    |  Winch'r  bu.,  2150.42  cu.  in.  Inip'l  bu.,  2218.192  cu.  in. 

Weight,         Troy  Ib.,  5760  gr.  Imperial  Ib.,  5760  gr.         Gramme,  15.433159  T.  gr. 

NOTES. — 1.  An  Imperial  gallon  is  equal  to  1.2  wine  gallons. 

2.  An  old  ale  or  beer  gallon  is  very  nearly  1.221  wine  gallons,  or  1.017  Im- 
perial gallons. 

3.  In  ordinary  computations  2150.4  cu.  in.  may  be  taken  as  a  Winchester 
bushel,  and  2218.2  cu.  in.  as  an  Imperial  bushel. 


MONEY  AND  CURRENCIES. 

343.  Money  is  the  commodity  adopted  to  serve  as  the  uni- 
versal equivalent  or  measure  of  value  of  all  other  commodities, 
and  for  which  individuals  readily  exchange  their  surplus  products 
or  their  services. 

344.  Coin  is  metal  struck,  stamped,  or  pressed  with  a  die,  to 
give   it  a  legal,  fixed  value,  for  the    purpose  of  circulating  as 
money. 

NOTE.  —  The  coins  of  civilized  nations  consist  of  gold,  silver,  copper,  and' 
nickel. 

345.  A  Mint  is  a  place  in  which  the  coin  of  a  country  or 
government  is  manufactured. 

NOTE.  —  In  all  civilized  countries  mints  and  coinage  are  under  the  exclusive 
direction  and  control  of  government. 


188  COMPOUND    NUMBERS. 

346.  An  Alloy  is  a   metal    compounded   with   another  of 
greater  value.     In  coinage,  the  less  valuable  or  baser  metal  is  not 
reckoned  of  any  value. 

NOTE. — "Gold  and  silver,  in  their  pure  state,  are  too  soft  and  flexible  for  coin- 
age; hence  they  are  hardened  by  compounding  them  with  an  alloy  of  baser 
metal,  while  their  color  and  other  valuable  qualities  are  not  materially  impaired. 

347.  An  Assayer  is  a  person  who  determines  the  composi- 
tion and  consequent  value  of  alloyed  gold  and  silver. 

The  fineness  of  gold  is  estimated  by  carats,  as  follows : — 

Any  mass  or  quantity  of  gold,  either  pure  or  alloyed,  is  divided 
into  24  equal  parts,  and  each  part  is  called  a  carat. 

Fine  gold  is  pure,  and  is  24  carats  fine. 

Alloyed  gold  is  as  many  carats  fine  as  it  contains  parts  in  24  of 
fine  or  pure  golfl.  Thus,  gold  20  carats  fine  contains  20  parts  or 
carats  of  fine  gold,  and  4  parts  or  carats  of  alloy. 

348.  An  Ingot  is  a  small  mass  or  bar  of  gold  or  silver,  in- 
tended either  for  coinage  or  exportation.     Ingots  for  exportation 
usually  have  the  assayer's  or  mint  value  stamped  upon  them. 

349.  Bullion  is  uncoined  gold  or  silver. 

3«IO.  Bank  Bills  or  Bank  Notes  are  bills  or  notes  issued  by 
a  banking  company,  and  are  payable  to  the  bearer  in  gold  or  silver, 
at  the  bank,  on  demand.  They  are  substitutes  for  coin,  but  are 
not  legal  tender  in  payment  of  debts  or  other  obligations. 

351.  Treasury  Notes  are  notes  issued  by  the  General  Govern- 
ment, and  are  payable  to  the  bearer  in  gold  or  silver,  at  the  gene- 
ral treasury,  at  a  specified  time. 

3*52.  Currency  is  coin,  bank  bills,  treasury  notes,  and  other 
substitutes  for  money,  employed  in  trade  and  commerce. 

353.  A  Circulating  Medium  is  the  currency  or  money  of  a 
country  or  government. 

354.  A  Decimal  Currency  is  a  currency  whose  denomina- 
tions increase  and  decrease  according  to  the  decimal  scale. 

I.  UNITED  STATES  MONEY. 

355.  The  currency  of  the  United  States  is  decimal  currency, 
and  is  sometimes  called  Federal  Money. 


MONEY  AND   CURRENCIES.  189 

The  unit  is  tlie  gold  dollar,  weighing  25.8  grains;  and  all  the  other 
denominations  are  either  divisors  or  multiples  of  this  unit. 

TABLE. 

10  mills  (m.)  make  1  cent ct. 

10  cents  "       1  dime d. 

10  dimes  "      1  dollar $. 

10  dollars  "      1  eagle E. 

I:NIT  EQUIVALENTS. 

ct.  m. 

d.                      1  =             10 

*.                  1     =         10  =         100 

]:.               1     =       10     =       100  =       1000 

1     =     10     =     100     =     1000  =     10000 

SCALE  —  uniformly  10. 

NOTKS.  — 1.   Federal  Money  was  adopted  by  Congress  in  1786. 
2.   The  character  $  is  supposed  to  be  a  contraction  of  U.  8.  (United  States),  the  U 
being  placed  upon  the  S. 

l>ythe  "Coinage  Act  of  1873,"  the  gold  coins  are  the  double 
eagle,  eagle,  half  eagle,  quarter  eagle,  three  dollar,  and  one  dollar 
pieces. 

The  silver  coins  are  the  trade  dollar,  the  half  dollar,  the  quarter 
dollar,  the  twenty-cent  and  the  ten-cent  pieces. 

The  nickel  coins  are  the  five-cent  and  three-cent  pieces. 

The  bronzf  coin  is  the  one-cent  piece. 

NOTES.  —  1.  The  trade  dollar  is  designed  solely  for  purposes  of  commerce,  and 
not  for  currency.  Its  weight  is  420  grains. 

2.  The  mill  is  a  denomination  used  only  in  computations :  it  is  not  a  coin. 

3*>G.  Government  Standard.  By  Act  of  Congress,  January 
18,  1837,  all  gold  and  silver  coins  must  consist  of  9  parts  (.900) 
pure  metal,  and  1  part  (.100)  alloy.  The  alloy  for  gold  must  con- 
sist of  equal  parts  of  silver  and  copper,  and  the  alloy  for  silver  of 
pure  copper. 

The  nickel  coins  are  75  parts  copper  and  25  parts  nickel. 

STATE    CURRENCIES. 

SoT.  United  States  money  is  reckoned  in  dollars,  dimes,  cents, 
and  mills,  o'he  dollar  being  uniformly  valued  in  all  the  States  at  ICO 
cents ;  but  in  many  of  the  States  money  was  formerly  reckoned  in 
dollars,  shillings,  and  pence. 


*190  COMPOUND   NUHBEKS. 

NOTE. — At  the  time  of  the  adoption  of  our  decimal  currency  by  Congress,  in  1786, 
the  cotonial  currency,  or  bills  of  credit,  issued  by  the  colonies,  had  depreciated  in  value, 
and  this  depreciation,  being  unequal  in  the  different  colonies,  gave  rise  to  the  different 
values  of  the  State  currencies.  This  usage,  however,  has  become  nearly,  if  not  quite, 
obsolete  all  over  the  country. 

Georgia  Currency. 
Georgia,  South  Carolina, $1  =  4s.  8d,  =  56d, 

Canada  Currency. 

The  Dominion  of  Canada, $1  =  5s.  =  60d. 

New  England  Currency. 

New  England  States,  Indiana,  Illinois,  ] 

Missouri,  Virginia,  Kentucky,  Tennes-  i- $1  =  6s.  =  72d. 

Bee,  Mississippi,  Texas, J 

Pennsylvania  Currency. 

New  Jersey,  Pennsylvania,  Delaware, )          A-J -     g^  _  QQ  •• 

Maryland, j  "  " 

New  York  Currency. 

New  York,  Ohio,  Michigan,  )  &i       o          O/M 

North  Carolina, . . ..  J |] 


II.     CANADA  MONET. 

358.  The  currency  of  the   Dominion  of  Canada  is  decimal, 
and  the  table  and  denominations  are  the  same  as  those  of  the  United 
States  money. 

NOTE.  — The  currency  of  the  whole  Dominion  of  Canada  was  made  uniform 
July  1,  1871.  Before  the  adoption  of  the  decimal  system,  pounds,  shillings  and 
pence  were  used. 

The  coin  of  Canada  is  of  silver  and  of  bronze. 

The  silver  coins  are  the  50-cent  piece,  25-cent  piece,  10-cent 
piece,  and  5-cent  piece.  The  20-cent  piece  is  no  longer  coined. 

The  bronze  coin  is  the  cent. 

The  gold  coin  used  in  Canada  is  the  British  sovereign,  worth 
$4.86§,  and  the  half  sovereign. 

The  intrinsic  value  of  the  50-cent  piece  in  United  States  money 
is  about  4G§  cents,  of  the  25-cent  piece  23  ?T  cents.  In  ordinary 
business  transactions  they  pass  the  same  as  United  Sta'tes  coin  of 
same  denomination. 

359.  Government  Standard.     The  silver  coins  consist  of  925 
parts  (.925)  pure  silver  and  75  parts  (.075)  copper.     That  is,  they 
are  .925  fine. 


MONEY  AND  CURRENCIES. 

III.    ENGLISH  MONET. 

36O.  English  or  Sterling  Money  is  the  currency  of  Great 
Britain. 

The  unit  is  the  pound  sterling,  and  all  the  other  denominations 
are  divisors  of  this  unit. 

TABLE. 

4  farthings  (far.  or  qr.)  make  1  penny, «.....»•«.....  ,d. 

12  pence  "       1  shilling, s. 

20  shillings  "      1  pound  or  sovereign  . .  £  or  sov. 

UNIT   EQUIVALENTS. 

d.  far. 

8.  1=4 

£,  or  SOT.   1  =     12  =    48 

1  =  20  =  240  =  960 
SCAEE  —  ascending,  4,  12,  20;  descending,  20,  12,  4. 

NOTES.  —  1.  Farthings  are  generally  expressed  as  fractions  of  a  penny;  thus, 

1  far.,  sometimes  called  1  quarter,  (qr.)=--|d.  j  3  far.  =  |d. 

2.  The  old/,  the  original  abbreviation  for  shillings,  was  formerly  written  be- 
tween shillings  and  pence,  and  d,  the  abbreviation  for  pence,  was  omitted.  Thus 
2s.  6d.  was  written  2/6.  A  straight  line  is  now  used  in  place  of  the/  and  shil- 
lings are  written  on  the  left  of  it  and  pence  on  the  right.  Thus,  2/6,  10/3, 
etc. 

COINS.  The  gold  coins  are  the  sovereign  (=  £1)  and  the  half 
sovereign,  (=  10s.) 

The  silver  coins  are  the  crown,  half  crown,  florin,  the  shilling, 
sixpenny,  fourpenny,  and  threepenny  pieces. 

The  copper  and  bronze  coins  are  the  penny,  half  penny,  and 
farthing. 

NOTE.  — The  guinea  (=  21s.)  and  the  half  guinea  (=  10a.  6d.  sterling)  are  old 
gold  coins,  and  are  no  longer  coined. 

3G1.  Government  Standard.  The  standard  fineness  of  Eng- 
lish gold  coin  is  11  parts  pure  gold  and  1  part  alloy;  that  is,  it  is 
22  carats  fine.  The  standard  fineness  of  silver  coin  is  11  oz. 

2  pwt.  (=  11.1  oz.)  pure  silver  to  18  pwt.  (=  .9  oz.)  alloy.    Hence 
the  silver  coins  are  11  oz.  2  pwt.  fine;  that  is,  11  oz.  2  pwt.  pure 
silver  in  1  Ib.  standard  silver. 

This  standard  is  37  parts  (|J  =  .925)  pure  silver  and  3  parts 
(^  =  .075)  copper. 

NOTE.  —  A  pound  of  English  standard  gold  is  equal  in  value  to  14.2878  Ib.  » 
14  Ib.  3  oz.  9  pwt.  1.727  gr.  of  silver. 


192  COMPOUND  NUMBERS. 

IV.     FRENCH  MONEY. 

362.     The  currency  of  France  is  decimal  currency. 
The  unit  is  the  franc,  of  -which  the  other  denominations  are  di- 
visors. 

TABLE. 

10  millimes  make  1  centime. 
100  centimes  "      1  franc. 

SCALK — ascending,  10,  100;  descending,  100,  10. 
C oils' s.     The  gold  coins  are  the  40,  20,  10,  and  5  franc  pieces. 
The  silver  coins  are  the  5,  2,  and  1  franc,  the  50  and  20  centime 
pieces. 

The  bronze  coins  are  the  10,  5,  2,  and  1  centime  pieces. 

363.       COMPARATIVE    TABLE    OF    MONEYS. 


English. 

U.  S.               French. 

US. 

1  penny, 

d. 

$0.0202  + 

1  centime, 

ct. 

$0.00193 

1  shilling, 

8. 

.2433  + 

1  decime, 

dc. 

0.0193 

1  florin, 

fl. 

.4866  + 

1  franc, 

fr. 

.193 

1  sovereign, 

sov. 

4.8665 

REDUCTION. 

364:.     Reduction  is  the  process  of  changing  a  number  from  one 
denomination  to  another  without  altering  its  value. 

Reduction  is  of  two  kinds,  Descending  and  Ascending. 

365.  Eeduction  Descending  is  changing  a  number  of  one  do- 
nomination  to  another  denomination  of  less  unit  value  ;   thus,  $1  = 
10  dimes  =  100  cents  =  1000  mills. 

366.  Eeduction  Ascending  is  changing  a  number  cf  cne  de- 
nomination to  another  denomination  of  greater   unit  value;   thus, 
1000  mills  =  100  cents  =  10  dimes  =  $1. 

REDUCTION    DESCENDING. 
CASE    I. 

367.  To  reduce  a  compound   number  to   lower  de- 
nominations. 

1.  Reduce  3  mi.  57  rd.  2  yd.  1  ft.  8  in.  to  inches. 


REDUCTION.  193 

OPERATION.  ANALYSIS.     Since 

3  mi.   57  rd.   2  yd.  1  ft.   8  in.  in  1  mile  there  are 

320  rd.,  in  3  miles 
there  are  3  x  320 
rd.  =  960  rd.,  and 
the  57  rd.  in  the 
given  number  add- 
508i  ed,  makes  1017  rd. 

5595$  yd.  in     3    mi.     57    rd. 

3  Since  in  1  rd.  there 


16787|  ft,  are5$yd.,in!017rd. 

12  there  are  1017  x  5$ 

2oT458~in.  ?<*•    -    6593$    yd., 

which  plus  the  2  yd. 

in  the  given  number  =  5595|  yd.  in  3  mi.  57  rd.  2  yd.  Since  in  1  yd. 
there  are  3  ft,,  in  5595$  yd.  there  are  5595|  x  3  ft.  =  16786$  ft., 
which  plus  the  1  ft.  in  the  given  number  =  16787$  ft,  in  3  mi.  57  rd. 
2  yd.  1  ft.  And  since  in  1  ft.  there  are  12  in.,  in  16787$  ft.  there  are 
16787$  x  12  in.  =  201450  in.,  which  plus  the  8  in.  in  the  given  num- 
ber =  201458  in.  in  the  given  compound  number.  On  examining  the 
operation,  we  find  that  we  have  successively  multiplied  by  the  num- 
bers in  the  descending  scale  of  linear  measure  from  miles  to  inches, 
inclusive.  But,  as  either  factor  may  be  used  as  a  multiplicand  (82, 1), 
we  may  consider  the  numbers  in  the  descending  scale  as  multipliers. 
Hence  the  following 

RULE.  T.  Multiply  the  highest  denomination  of  the  given 
compound  number  by  that  number  of  the  scale  which  will  reduce  it 
to  the  next  lower  denomination,  and  add  to  the  product  the  given 
number,  if  any,  of  that  lower  denomination. 

II.  Proceed  in  the  same  manner  with  the  results  obtained  in  each 
lower  denomination,  until  the  reduction  is  brought  to  the  denomina- 
tion required. 

EXAMPLES    FOR    PRACTICE. 

1.  In  16  lb.  10  oz.  18  pwt.  5  gr.,  how  many  grains? 
17 


194  COMPOUND  NUMBERS. 

i 

2.  In  £133  6  s.  8d.,  how  many  farthings?     Am.  128,000. 

3.  Change  100  mi.  to  inches.  Ans.  G33GOOO  in. 

4.  How  many  rods    of  fence  will   inclose  a  farm   11    miles 
square?  Ans.  1920  rd. 

5.  The  grey  limestone  of  Central  New  York  weighs  175  Ibs. 
to  the  cubic  foot;  what  is  the  weight  of  a  block  8  ft.  long  and 
1  yd.  square  ?  Ans.  6  T.  6  cwt. 

3.  What  will  be  the  cost  of  1  hhd.  of  molasses  at  $.28  per  gal.  ? 

7.  A  man  wishes  to  ship  1548  bu.  1  pk.  of  potatoes  in  barrels 
containing  2  bu.  3  pk.  each ;  how  many  barrels  must  he  obtain  ? 

8.  A  grocer  bought  10  bu.  of  chestnuts  at  $3.75  a  bushel,  and 
retailed  them  at  $.06£  a  pint;  how  much  was  his  whole  gain  ? 

9.  Reduce  90°  17'  40"  to  seconds.  Ans.  325060". 

10.  In  the  18th  century  how  many  days?      Ans.  36524  da. 

11.  At  6^  cts.  each,  what  will  be  the  cost  of  a  great-gross  of 
Writing  books  ?  Ans.  $108. 

12.  How  large  an  edition  of  an  octavo  book  can  be  printed 
from  4 .  bales  4  bundles  1  ream  10  quires  of  paper,  allowing  8 
sheets  to  the  volume  ?  Ans.  2970  vol. 

13.  Suppose  your  age  to  be  18  yr.  24  da.  ;  how  many  minutes 
old  are  you,  allowing  4  leap  years  to  have  occurred  in  that  time  ? 

14.  How  many  pence  in  481  sovereigns?      Ans.  115,440  d. 

15.  Reduce  $7f  to  mills.  Ans.  7375  mills. 

16.  In  3  P.  of  Sherry  wine,  how  many  qt.  ?      Ans,  1560  qt. 

17.  Reduce  37  Eng.  ells  1  qr.  to  yd.         Ans.  46  yd.  2  qr. 

18.  In  £6  10s.  lOd.  how  many  dollars  U.  S.  currency  ? 

19.  Reduce  6  0.  14fg  3f£  45iT[  to  minims. 

20.  Reduce  1  T.  1  P.  1  hhd.  to  Imperial  gallons. 

Ans.  367  i  Imperial  gal. 

21.  How  many  dollars  Canada  currency  arc  equal  to  £126  12s. 
6d.?  Ans.  $500*. 

22.  ITow  many  pint,  quart,  and  two-quart  bottles,  of  each  an 
equal  number,  may  be  rilled  from  a  hogshead  of  wine  ? 

Ans.  72. 

23.  How  many  steps  of  2  ft.  9  in.  each,  will   a   man  take,  ia 
walking  from  Erie  to  Cleveland,  tho  distance  being  95  mi.? 


KEDUCTION.  195 

24.  A  grocer  bought  12  bbl.  of  cider  at  $lf  a  barrel,  and  after 
converting  it  into  vinegar,  he  retailed  it  at  6  cents  a  quart ;  how 
much  was  his  whole  gain  ?  Ans.  $69.72. 

25.  In  75  A.  4  sq.  ch.  18  P.  118  sq.  1.  how  many  square  links? 

26.  How  many  inches  high  is  a  horse  that  measures  16  hands? 

27.  If  a  vessel  sail  150  leagues  in  a  day,  how  many  statute 
miles  does  she  sail  ?  Ans.  517.5. 

1  28.  If  14  A.  be  sold  from  a  field  containing  50  A.,  how  many 
square  rods  will  the  remainder  contain  ?  Ans.  5,760  sq.  rd. 

29.  A  man  returning  from  Pike's  Peak  has  36  Ib.  8  oz.  of 
pure  gold ;  what  is  its  value  at  $1.04 £  per  pwt.  ?  Ans.  §9169.60. 

80.  A  person  having  8  hhd.  of  tobacco,  each  weighing  9  cwt. 
42  Ib.,  wishes  to  put  it  into  boxes  containing  48  Ib.  each ;  how 
many  boxes  must  he  obtain  ?  Ans.  157. 

31.  A  merchant  bought  12  bbl.  of  salt  at  $1J  a  barrel,  and  re- 
tailed it  at  f  of  a  cent  a  pound;  how  much  was  his  whole  gain? 

32.  A  physician  bought  lib  10^  of  quinine  at  $2.25  an  ounce, 
and  dealt  it  out  in  doses  of  10  gr.  at  $.12£  each;  how  much  more 
than  cost  did  he  receive  ?  Ans.  $82.50. 


CASE   II. 

368.  To  reduce  a  denominate  fraction  from  a  greater 
to  a  less  unit. 

1.  Reduce  ^  of  a  gallon  to  the  fraction  of  a  gill. 

OPERATION.  ANALYSIS.      To  re- 

A  gal.  X  f  x  f  X  f  =  T8T  gi.  duce  gallons  to  gills, 

~  we    multiply    succes- 

sively by  4,  2,  and  4, 

1  the  numbers  in  the  de- 
scending scale.     And 

2  since  the  given  num- 
J2 ber  is  a  fraction,  we 

11  8  =  T8T  gi.,  Ans.  indicate  the  process, 
as  in  multiplication  of 
fractions,  after  which  we  perform  the  indicated  operations,  and  ob- 
tain -j^j,  the  answer.  Hence, 


196  COMPOUND  NUMBERS. 

RULE.  Multiply  the  fraction  of  the  higher  denomination  l>y  the 
numbers  in  the  descending  scale  successively,  between  the  given  and 
the  required  denomination. 

NOTE.  —  Cancellation  may  be  applied  wherever  practicable. 

EXAMPLES    FOR    PRACTICE. 

1.  Reduce  ^^  of  a  Ib.  Troy  to  the  fraction  of  a  pennyweight. 

Ans.   |  pwt. 

2.  Reduce  g|2  of  a  hhd.  to  the  fraction  of  a  pint. 

3.  Reduce  3773  of  a  mile  to  the  fraction  of  a  yard. 

Ans.    |  yd. 

4.  Reduce  ^|2  of  a  gallon  to  the  fraction  of  a  gill. 

5.  What  part  of  an  ounce  is-g-g^  of  f  of  f  of  T\  of  3-j[-  pounds 
avoirdupois  weight  ?  Jy/.s-.  ^jjj-f-g-  oz. 

6.  Reduce         --   of  a  dollar  to  the  fraction  of  a  cent. 


7.  Reduce  ^  of  a  rod  to  the  fraction  of  a  link.     Ans.  |  1. 

8.  Reduce  ^U  of  a  scruple  to  the  fraction  of  a  grain. 

9.  What  fraction  of  a  yard  is  f  of  y*T  of  a  rod  ? 

10.  -j65  of  a  week  is  |  of  how  many  days?  Ans.  8|  da. 

11.  What  fraction  of  a  square  rod  is  yg3^  of  4|  times  T2^  of  an 
acre  ?  Aits.  T3^  sq.  rd. 

CASE   ITT. 

369.  To  reduce  a  denominate  fraction  to  integers 
of  lower  denominations. 

1.  What  is  the  value  of  f  of  a  bushel  ? 

OPERATION.  ANALYSIS,     f  bu.  =  f  of 

I  bu.  X  4  =    f  pk.  =  If  pk.  4  pk.,  or  lg  pk.  ;   2  pk.  =  2 

3  pk    x  8  =  ^  qt.  =  44  qt.  °f  8  qt.  =  44  qt.  ;  and  |  qt. 

4qt,    X2=    f   pt.  =  Hpt.  =jof2pt.  =  lipt.    The 

1  pk.  4  qt.  1|  pt.,  Ans.  Uni*8'  I   Pk-'  4/"  \  ^> 

with    the    last  denominate 

fraction,  f  pt.,  form  the  answer.     Hence, 

RULE.  I.  Multiply  the  fraction  by  that  number  in  the  scale 
which  will  reduce  it  to  the  next  lower  denomination,  and  if  the 
result  be  an  improper  fraction,  reduce  it  to  a  whole  or  mixed 
number. 


REDUCTION.  197 

IT.  Proceed  with  the  fractional  part,  if  any,  as  before,  until 
reduced  to  the  denominations  required. 

III.  The  units  of  the  several  denominations)  arranged  in  their 
order,  will  be  the  required  result. 

EXAMPLES   FOR   PRACTICE. 

1.  Reduce  T9^  of  a  yard  to  integers  of  lower  denominations. 

Ans.  2  ft.  8 1  in. 

2.  Reduce  j  of  a  month  to  lower  denominations. 

3.  Reduce  {jj  J  of  a  short  ton  to  lower  denominations. 

4.  What  is  the  value  of  J  of  a  long  ton  ? 

Ans.  11  cwt.  12  Ib.  7^  oz. 

5.  What  is  the  value  of  |  of  2^  pounds  apothecaries'  weight? 

6.  What  is  the  value  of  -?%  of  an  acre  ? 

Ans.  86  P.  4  sq.  yd.  5  sq.  ft.  127T53  sq.  in. 

7.  Reduce  |  of  a  mile  to  integers  of  lower  denominations. 

8.  What  is  the  value  of  ^  of  a  great  gross? 

Ans.  6  gross  10  doz.  3f . 

9.  What  is  the  value  in  geographic  miles  of  T9ff  of  a  great 
circle?  Ans.  12150  mi. 

10.  What  is  the  value  of  |  of  3|  cords  of  wood  ? 

Am.  2  Cd.  5  cd.  ft.  9|  cu.  ft. 

11.  The  distance  from  Buffalo  to  Cincinnati  is  438  miles;  hav- 
ing traveled  |  of  this  distance,  how  far  have  I  yet  to  travel  ? 

Ans.  262  mi.  256  rd. 

12.  What  is  the  value  of  ||  f^  ?  .4ns.  3  f^  35  T^. 

13.  What  is  the  value  of  f  of  a  sign? 

Ans.  12°  51'  25f". 

14.  A  man  having  a  hogshead  of  wine,    sold  T63  of  it;  how 
much  remained  ?  Ans.  33  gal.  3  qt.  1  pt.  !T7g  gi. 


CASE    IV. 

3 TO.  To  reduce  a  denominate  decimal-  to  integers 
of  lower  denominations. 

1.  Reduce  .125  of  a  barrel  to  integers  of  lower  denominations. 
17* 


198  COMPOUND  NUMBERS. 

i 

OPERATION.  ANALYSIS.      We   first  multiply 

125  the  given  decimal,  .125  of  a  barrel, 

81.5  by  31.5  (=  31J)   to  reduce  it  to 

o  no--       ,  gallons,  and  obtain  3.9375  gallons. 

Omitting  the  3  gallons,  AVC  mul- 

tiply the  decimal,  .9375  gal.,  by 

3.  /  500  qt.  4  to  reduce  it  to  quarts,  and  obtain 

3.75    quarts.     We    next   multiply 

1.50  pt.  the  decimal  part  of  this  result  by 

4  2,  to  reduce  it  to  pints,  and  obtain 

2~Q  „{  1.5  pints.     And  the  decimal  part 


3  gal.  3  qt.  lpt.  2  gi.,  Am.      of  this  r.esult  we  multiP^  b?  4  to 

reduce   it  to  gills,   and  obtain  2 

gills.  The  integers  of  the  several  denominations,  arranged  in  their 
order,  form  the  answer.  Hence, 

RULE.  I,  Multiply  the  given  denominate  decimal  ly  that  num- 
ber in  the  descending  scale  which  will  reduce  it  to  the  next  lower 
denomination,  and  point  off  the  result  as  in  multiplication,  of 
decimals. 

II.  Proceed  with  the  decimal  part  of  the,  product  in  the  same 
manner  until  reduced  to  the  required  denominations.  The  integers 
at  the  left  will  be  the  answer  required. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  value  of  .645  of  a  day  ? 

Ans.  15  h.  28  min.  48  sec. 

2.  What  is  the  value  of  .765  of  a  pound  Troy  ? 

3.  What  is  the  value  of  .6625  of  a  mile? 

4.  WThat  is  the  value  of  .8469  of  a  degree  ? 

Ans.  50'  48.84". 

5.  What  is  the  value  of  .875  of  a  hhd.  ? 

6.  What  is  the  value  of  £.85251  ? 

Ans.  17  s.  2.4  -f  far. 

7.  What  is  the  value  of  .715°?  Ans.  42'  54". 

8.  What  is  the  value  of  7.88125  acres? 

Ans.  7  A.  141  P. 

9.  What  is  the  value  of  .625  of  a  fathom  ?  Ans.  3|  ft. 


REDUCTION.  199 

10.  What  is  the  value  of  .375625  of  a  barrel  of  pork? 

11.  What  is  the  value  of  .1150390625  Cong.  ? 

Ans.  14  f 5  5f^48  m. 

12.  What  is  the  value  of  .61  of  a  tun  of  wine  ? 

Ans.  I  P.  27  gal.  2  qt.  1  pt.  3.04  gi. 

REDUCTION   ASCENDING. 
CASE    I. 

371.    To  reduce  a  denominate  number  to  a  com- 
pound number  of  higher  denominations. 

1.  Reduce  157540  minutes  to  weeks. 

ANALYSIS.       Dividing 
OPERATION. 

the     given     number    of 

60).!57540  min-  minutes   by  CO,  because 

24  )  2625  h.  -f-  40  min.  there    are   Js   as    many 

7  ^  1 00   d     4-  0  h  hours  as  minutes,  and  we 

obtain  2G45  h.  plus  a  re- 

15  wk.  -f  4  da.  mainder  of  40  min.     Vie 

15  wk.  4  da.  9  h.  40  min.,  A?is.         ncxt  divide  the  2G45  h. 

by  24,  because  there  are 

.Jj  as  many  days  as  hours,  and  we  find  that  2G45  h.  =  109  da.  plus  a 
remainder  of  9  h.  Lastly  we  divide  the  109  da.  by  7,  because  there 
are  \-  as  many  weeks  as  days,  and  we  find  that  109  da.  =  15  wk.  plus 
a  remainder  of  4  da.  The  last  quotient  and  the  several  remainders 
annexed  in  the  order  of  the  succeeding  denominations,  form  the 
answer. 

2.  Reduce  201458  inches  to  miles. 

ANALYSIS.      We 

OPERATION.  divide    successively 

12)201458  in.  by  the   numbers  in 

3)16788  ft.  2  in.  tlie  ascending  scale 

of   linear    measure, 
5J  or  5.5)5596  yd.  in  tfae  game  manner 

320)1017  rd.  2  yd.  1  ft.  6  in.  as  in  the  last  pre- 

~3  mi,  57  rd.  cedinS      operation. 

But,  in  dividing  the 

3  mi.  57  rd.  2  yd.  1  ft,  8  in.,  Ans.  5596  yd.   by  5£  or 

5.5.  we  have  a  re- 


200  COMPOUND  NUMBERS. 

( 

mainder  of  2^  yd.,  and  this  reduced  to  its  equivalent  compound  num- 
ber, (369)  =  2  yd.  1  ft.  6  in.  In  forming  our  final  result,  the  6  in. 
of  this  number  are  added'  to  the  first  remainder,  2  in.,  making  the  8  in. 
as  given  in  the  answer.  From  these  examples  and  analyses  we  deduce 
the  following 

RULE.  I.  Divide  the  given  concrete  or  denominate  number  by 
that  number  of  the  ascending  scale  which  will  reduce  it  to  the  next 
higher  denomination. 

II.  Divide  the  quotient  by  the  next  higher  number  in  the  scale  / 
and  so  proceed  to  the  highest  denomination  required.  The  last 
quotient,  with  the  several  remainders  annexed  in  a  reversed  order, 
will  be  the  answer. 

NOTE.  —  The  several  corresponding  cases  in  reduction  descending  and  reduc- 
tion ascending,  being  opposites,  mutually  prove  each  other. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  1913551  ounces  to  tons. 

2.  In  97920  gr.  of  medicine  how  many  Ib.  ?        Ans.  17  Ib. 

3.  Reduce  1000000  in.  to  mi. 

4.  How  many  acres  in  a  field  120  rd.  long  and  56  rd.  wide  ? 

5.  In  a  pile  of  wood  60  ft.  long,  15  ft.  wide,  and  10  ft.  high, 
how  many  cords  ?  Ans.  70  Cd.  2  cd.  ft.  8  cu.  ft. 

6.  How  many  fathoms  deep  is  a  pond  that  measures  28  ft.  6 
in.  ?  Ans.  4|  fath. 

7.  In  30876  gi.  how  many  hhd.  ? 

8.  How  many  bushels  of  corn  in  27072  qt.  ?     Ans.  846  bu. 

9.  At  2  cts.  a  gill,  how  much  alcohol  may  be  bought  for  $'2.54  ? 

10.  In  1234567  far.  how  many  £?  Ans.  £1286  If  d. 

11.  Reduce  2468  pence  to  half  crowns. 

12.  In  $88.35  how  many  francs?  Ans.  475. 

13.  In  622080  cu.  in.  how  many  tons  of  round  timber  ? 

14.  In  84621  ty  how  many  Cong.  ? 

15.  If  135  million  Gillott  steel  pens  are  manufactured  yearly^ 
how  many  great-gross  will  they  make?  Ans.  78125. 

16.  Reduce  1020300"  to  S.  Ans.  9  S.  13°  25' 

17.  In  411405  sec.  how  many  da.  ? 


REDUCTION.  201 

18.  During  a  storm  at  sea,  a  ship  changed   her  latitude  412 
geographic  miles  ;  how  many  degrees  and  minutes  did  she  change  ? 

An*.  6°  52'. 

19.  If  a  man  travel  at  the  rate  of  a  minute  of  distance  in  20 
minutes  of  time,  how  much  time  would  he  require  to  travel  round 
the  earth?  Ans.  300  days. 

20.  In  120  gross  how  many  score  ?  Ans.  864. 

21.  How  many  miles  in  the  semi-circumference  of  the  earth? 

22.  How  much  time  will  a  person  gain  in  36  yr.  by  rising  45 
min.  earlier,  and  retiring  25  ruin,  later,  every  day,  allowing  for  9 
leap  years  ?  Ans.  639  da.  4  h.  30  niin. 

23.  A  grocer  bought  20  gal.  of  milk  by  beer  measure,  and  sold 
it  by  wine  measure ;  how  many  quarts  did  he  gain  ?     Ans.  17|4- 

24.  How  many  bushels  of  oats  in  Connecticut  are  equivalent  to 
1500  bushels  in  Iowa  ?  Ans.  1875  bu. 

25.  Reduce  120  leagues  to  statute  miles.        Ans.  414.96  mi. 

26.  In  1  bbl.  1  gal.   2  qt.  wine  measure,  how  many  beer  gal- 
lons ?  Ans.  27^\. 

27.  Reduce  150  U.  S.  bushels  to  Imperial  bushels. 

Ans.  145.415  +  Imp'l.  bu. 

28.  How  many  squares  in  a  floor  68  ft.  8  in.  long,  and  33  ft. 
wide?  Ans.  22fjj. 

29.  How  many  cubic  inches  in  a  solid  4  ft.  long  3  ft.  wide,  and 
1  ft.  6  in.  thick  ? 

30.  How  many  acres  in  a  field  120  rd.  long  and  56  rd.  wide  ? 

31.  Change  356  dr.  apothecaries  weight,  to  Troy  weight. 

32.  A  coal  dealer  bought  175  tons  of  coal  at  $3.75  by  the  long 
ton,  and  sold  it  at  $4.50  by  the   short  ton;  how  much  was  his 
whole  gain  ?  Ans.  $225.75. 

33.  How  many  acres  of  land  can  be  purchased  in  the  city  of 
iNew  York  for  $73750,  at  $1.25  a  square  foot? 

Am.  1  A.  56  P.  194  sq.  ft. 

34.  An  Ohio  farmer  sold  a  load  of  corn  weighing  2492  lb.,  and  a 
load  of  wheat  weighing  2175  lb. ;  for  the  corn  he  received  $.60  a 
bushel,  and  for  the  wheat  $1.20  a  bushel;  how  much  did  he  re- 
ceive for  both  loads  ?  Ans.  $70.20. 


202 


COMPOUND  NUMBERS. 


The  following  examples  are  given  to  illustrate  a  short  and  prac- 


tical method  of  reducing  currencies. 


35.  What  will  be  the  cost  of  54  bu.  of  corn  at  5s.  a  bushel, 
New  England  currency  ? 

OPERATION. 


Or, 


54  x  5 

270s.  -j-  6 


270s. 
$45 


9         ANALYSIS.    Since  1  bu.  costs 

{       5s.,54bu.  cost54x5s.  =  270s.; 

and  since  Gs.  make  $1  N.  E. 

:-=.      currency,  270  ~-  6  =  $45,  Ans. 


OPERATION. 


36.  What  will  270  bu.  of  wheat  cost,  @  8s.  4d.  Penn.  currency  ? 

ANALYSIS.  Multiply  the 
quantity  by  the  price  in  Penn. 
currency,  and  divide  the  cost 
by  the  value  of  $1  in  the  same 
currency;  or  reduce  the  shil- 
lings and  pence  to  a  fraction 


100 


Or, 


25 

2 

POO 


of  a  shilling,  before  multiply- 
ing and  dividing. 

37.  Bought  5  hhd.  of  rum  at  the  rate  of  2s.  4d.  a  quart,  Geor- 
gia currency ;  how  much  was  the  whole  cost  ? 

OPERATION. 

5  5  ANALYSIS.      In   this   ex- 

63  63  ample   we    first    reduce    5 

Or,  2  hhd.  to  quarts  by  multiply- 

^  ing  by  63  and  4,  and  then 

F  proceed  as  in  the  preceding 

examples. 


$630 


$630 


38.  Sold  120  barrels  of  apples,  each  containing  2  bu.  2  pk.,  at 
4s.  7d.  a  bushel,  and  received  pay  in  cloth  at  10s.  5d.  a  yard ;  how 
many  yards  of  cloth  did  I  receive  ? 

OPERATION.  ANALYSIS.     The  operation  in  this  example  is 

12          similar  to  the  preceding  examples,  except  that  we 
440  divide  the  cost  of  the  apples  by  the  price  of  a  unit 

of  the  article  received  in  payment,  reduced  to  units 
of  the  same  denomination  as  the  price  of  a  unit  of 
the  article  sold.     The  result  will  be  the  same  in 
$132  whatever  currency. 


11 


tt 


REDUCTION.  203 

39.  What  cost  75  yards  of  flannel  at  3s.  6d.  per  yard.  New 
England  currency  ?  Ans.  $43.75. 

40.  A  man  in  Philadelphia  worked  5  weeks  at  6s.  4d.  a  day  ; 
how  much  did  his  wages  amount  to  ?  A»s.  $25.83£. 

41.  A  farmer  exchanged  2  bushels  of  beans  worth  10s.  Cd.  per 
bushel,  for  two  kinds  of  sugar,  the  one  at  lOd.  and  the  other  at 
lid.   per  pound,  taking  the  same  quantity  of  each  kind;  how 
many  pounds  of  sugar  did  he  receive?  Ans.  24  Ib. 

42.  If  corn  be  rated  at  5s.  lOd.  per  bushel  in  Vermont,  at  what 
price  in  the  currency  of  New  Jersey  must  it  be  sold,  in  order  to 
gain  $7.50  on  54  bushels? 

CASE    II. 

372.  To  reduce  a  denominate  fraction  from  a  less  to 
a  greater  unit. 

1.  Reduce  T8r  of  a  gill  to  the  fraction  of  a  gallon. 

OPERATION.  ANALYSIS.     To   re- 


T8T  S1-  *  i  X  £  X  -  i  =  ,V  gal.  ducc  Sllls  to 

we  divide  successively 

Or>  by   4,   2,   and  4,  the 


11 
4 

i 

44 


numbers  in  the  as- 
cending scale.  And 
since  the  given  num- 
ber  is  a  fraction,  we 


1  =  ^j,  Ans.  indicate  the  process, 

as  in  division  of  frac- 
tions, after  which  we  perform  the  indicated  operations,  and  obtain 
i*f ,  the  answer.  Hence, 

RULE.  Divide  the  fraction  of  the  lower  denomination  by  the 
numbers  in  the  ascending  scale  successively,  between  the  given  and 
the  required  denomination. 

NOTE. — The  operation  mny  frequently  be  shortened  by  cancellation. 

EXAMPLES    FOR   PRACTICE. 

1.  Keduce  f  of  a  shilling  to  the  fraction  of  a  pound. 

Ans. 


204  COMPOUND  NUMBERS. 

2.  Reduce  |  of  a  pennyweight  to  the  fraction  of  a  pound 
Troy.  An,.^\\>. 

3.  What  part  of  a  ton  is  j  of  a  pound  avoirdupois  weight  ? 

4.  What  fraction  of  an  hour  is  ^  of  20  seconds  ? 

5.  What  is  the  fractional  difference  between   g|^  of  a  hhd. 
and  |  of  a  pt.  ?      .  Ans.   35'^  hhd. 

6.  2|4  of  l  of  §  of  a  pint  is  what  fraction  of  2  pecks  ? 

Ans.   | 

7.  Reduce  f  of  |   of  T\  of  a  cord  foot  to  the  fraction  of  a 
cord.  Ans.  -£$  Cd. 

8.  What  part  of  an  acre  is  T3^  of  T47  of  9^  square  rods  ? 

9.  |-  of  220  rods  is  \  of  ^  of  how  many  miles? 

Ans.  12-J  mi. 

10.  A  block  of  granite  containing  |  of  ^  of  20  J  cubic  feet,  is 
what  fraction  of  a  perch?  Ans.  -Ji  Pch. 

11.  What  part  of  a  cord  of  wood  is  a  pile  7^  ft.  long,  2  ft.  high, 
and  3|  feet  wide  ?  Ans.  iff  Cd. 

12.  Reduce  f  of  an  inch  to  the  fraction  of  a  yard. 

CASE    III. 

373.  To  reduce  a  compound  number  to  a  fraction  of 
a  higher  denomination. 

1.  Reduce  2  oz.  12  pwt.  12  gr.  to  the  fraction  of  a  pound  Troy. 

OPERATION.  ANALYSIS.      To  find 

2  oz.  12  pwt.  12  gr.  =  1260  gr.  what  part  one  compound 

1  Ib.  Troy  —  5760  gr.  number   is   of  another, 

if  IS  lb-  =  372  lb->  Ans-       they  must  be  like  num* 

bers  and  reduced  to  the 

same  denomination.  In  2  oz.  12  pwt.  12  gr.  there  are  1260  gr.,  and 
in  1  Ib.  there  are  5760  gr.  Therefore  1  gr.  ia  3^ff  Ib.,  and  1260  gr. 
are  £*££  Ib.  =  ^  Ib.,  the  answer.  Hence, 

RULE.  Reduce  the  given  number  to  its  lowest  denomination  for 
the  numerator,  and  a  unit  of  the  required  denomination  to  the  same 
denomination  for  the  denominator  of  the  required  fraction. 

NOTE. — If  the  given  number  contain  a  fraction,  the  denominator  of  this  frac- 
tion must  be  regarded  as  the  lowest  denomination. 


REDUCTION.  205 

EXAMPLES   FOR   PRACTICE. 

1.  Reduce  100  P.  to  the  fraction  of  an  acre. 

Ans.  •{•  A. 

2.  What  part  of  a  mile  is  266  rd.  3  yd.  2  ft.  ? 

3.  What  part  of  a  £  is  18s.  5/1.  2-^  far.  ?  Ans.  £{f . 

4.  What  part  of  21  Ib.  Apothecaries'  weight  is  7^  7  3  23  14 
gr.  ?  Ans.  25^  Ib. 

5.  What  part  of  3  weeks  is  4  da.  16  h.  30  min.  ? 

6.  Reduce  1|  pecks  to  the  fraction  of  a  bushel. 

7.  From  a  hogshead  of  molasses  28  gal.  2  qt.  were  drawn; 
what  part  of  the  whole  remained  in  the  hogshead  ?       Ans.  |4- 

8.  Reduce  4  bundles  6  quires  16  sheets  of  paper  to  the  frac- 
tion of  a  bale.  Ans.   §  of  a  bale. 

9.  What  part  of  54  cords  of  wood  is  4800  cubic  feet  ? 

10.  What  is  the  value  of  ^  of  a  dollar  1  Ant.  36.30. 

11.  Reduce  3O.  3f 3  If  3  3 6  Iff.  to  the  fraction  of  a  Cong. 

12.  What  part  of  a  ton  of  hewn  timber  is  36  cu  ft.  864  cu.  in.  ? 

CASE   IV. 

374.  To  reduce  a  compound  number  to  a  decimal  of 
a  higher  denomination. 

1.  Reduce  3  cd.  ft.  8  cu.  ft.  to  the  decimal  of  a  cord. 

OPERATION.  ANALYSIS.     We  re- 

16     8.0         cu.  ft.  duce  the  8  cu.  ft.  to 

the  decimal  of  a  cd. 
ft.,  by  annexing  a  ci- 

.4375  Cd.,  Ans.  pher,    and    dividing 

Or,  by  16,  the  number  of 

3  cd.  ft.  8  cu.  ft.  =  56  cu.  ft.  cu.  ft.  in  1  cd.  ft.,  an- 

1  Cd.      a=  128  cu.  ft.  nexing    the    decimal 

jy5g  Cd.  =  T7g  Cd.  =  .4375  Cd.,  Ans.         quotient  to  the  3  cd. 

ft.     We   now  reduce 

the  3.  5  cd.  ft.  to  Cd.  or  a  decimal  of  a  Cd.,  by  dividing  by  8,  the 
number  of  cd.  ft.  in  1  Cd.,  and  we  have  .4375  Cd.,  the  answer. 
Or,  we  may  reduce  the  3  cd.  ft.  8  cu.  ft.,  to  the  fraction  of  a  Cd., 
18 


20C  COMPOUND  NUMBERS. 

(as  in  373),  and  we  shall  have  TVg  Cd.  =  T7F  Cd.,  which,  reduced  to 
its  equivalent  decimal,  equals  .4375  Cd.,  the  same  as  before.  Hence, 

RULE.  Divide  the  lowest  denomination  given  by  that  number 
in  the  scale  which  will  reduce  it  to  the  next  higher  denomination, 
and  annex  the  quotient  as  a  decimal  to  that  higher.  Proceed  in 
the  same  manner  until  the  whole  is  reduced  to  the  denomination 
required.  Or, 

Reduce  the  given  number  to  a  fraction  of  the  required  denomi- 
nation, and  reduce  this  fraction  to  a  decimal. 

EXAMPLES    FOR   PRACTICE. 

1.  Reduce  5  da.  9  h.  46  min.  48  sec.  to  the  decimal  of  a  week. 

Ans.  .7725  wk. 

2.  Reduce  3°  27'  46.44"  to  the  decimal  of  a  sign. 

3.  Reduce  51.52  P.  to  the  decimal  of  an  acre. 

4.  What  part  of  4  oz.  is  2  oz.  1G  pwt.  19.2  gr.  ?      Ans.  .71. 

5.  What  part  of  a  mile  is  28  rd.  2  yd.  1  ft.  11.04  in.  ? 

6.  Reduce  3J  |  to  the  decimal  of  a  pound. 

7.  Reduce  126  A.  4  sq.  ch.  12  P.  to  the  decimal  of  a  town- 
ship. Ans.  .0054893+  Tp. 

8.  What  part  of  a  fathom  is  3f  ft,  ?  Ans.  .625  fath. 

9.  What  part  of  1J  bushels  is  .45  of  a  peck?  Ans.  .09, 

10.  What  part  of  3  A.  80  P.  is  51.52  P.  ?  Ans.  .092. 

11.  Reduce  f  of  \  of  22|  Ib.  to  the  decimal  of  a  short  ton. 

12.  What  part  of  a  f%  is  5  f^  36  Tit  ?  Ans.  .1  fg. 

13.  Reduce  50  gal.  3  qt.  1  pt.  to  the  decimal  of  a  tun. 

Am.  .20188  -f  T. 
ADDITION. 

37*5.  Compound  numbers  are  added,  subtracted,  multiplied, 
and  divided  by  the  same  general  methods  as  are  employed  in 
simple  numbers.  The  corresponding  processes  are  based  upon  the 
same  principles;  and  the  only  modification  of  the  operations  and 
rules  is  that  required  for  borrowing,  carrying,  and  reducing  by  a 
varying,  instead  of  a  uniform  scale. 

876.  1.  What  is  the  sum  of  50  hhd.  32  gal.  3  qt.  1  pt.,  2 
hhd.  19  gal  1  pt,  15  hhd.  46J  gal.,  and  9  hhd.  39  gal.  2£  qt.? 


ADDITION.  207 

OPERATION.  ANALYSIS.     Writing  the  numbers  so  that 

hhd.    gal.     qt.    pt.         units  of  the  same  denomination  shall  stand 

50     32     31          in  the  same  column,  we  add  the  numbers 

of  the  right  ^and  or  lowest  denomination, 

and  find  the  amount  to  be  3  pints,  which  is 

J? ^     2     *          equal  to  1  qt.  1  pt.  We  write  the  1  pt.  under 

78      11     3     1          the  column  of  pints,  and  add  the  1  qt.  to 
the  column  of  quarts.     The  amount  of  the 

numbers  of  the  next  higher  denomination  is  7  qt.,  which  is  equal  to 
i  gal.  3  qt.  We  write  the  3  qt.  under  the  column  of  quarts,  and  add 
"he  1  gal.  to  the  column  of  gallons.  Adding  the  gallons,  we  find  the 
amount  to  be  137  gal.,  equal  to  2  hhd.  11  gal.  Writing  the  11  gal. 
under  the  gallons  in  the  given  numbers,  we  add  the  2  hhd.  to  the 
column  of  hogsheads.  Adding  the  hogsheads,  we  find  the  amount  to 
be  78  hhd.,  which  we  write  under  the  left  hand  denomination,  as  in 
simple  numbers. 

2.  What  is  the  sum  of  ^  wk.,  f  da.,  and  |  h. 

OPERATION.  ANALYSIS.       We 

T75  wk.  =  4  da.  21  h.  36  min.  first  find  the  value 

|   da.   =  14  "  24  min.  of  each  fraction  in 

|   h.    =                      22    «    30  sec.         inte?ers  °f  less  de~ 
-* nominations,  (369), 

5  and   then    add   the 

Or,  resulting  or  equiva- 

|   da,    X  ^  =  33-  wk;  lent  compound  num- 

|    h.  X  ^  X  ^  =  -g^  wk ;  bers. 

/5  wk.  -f  &  wk.  -f  ^  wk.  =  { *  wk ;  Or,    we   may   re- 

|  J  wk.  =  5  da.  12  h.  22  min.  30  sec.         duce  the  Siven  frac' 

tions  to  fractions  of 

the  same  denomination,  (368  or  372),  then  add  them,  and  find  the 
value  of  their  sum  in  lower  denominations. 

377.  From  these  examples  and  illustrations  we  derive  the 
following 

RULE.  I.  If  any  of  the  numbers  are  denominate  fractions,  or 
if  any  of  the  denominations  are  mixed  numbers,  reduce  the  frac- 
tions to  inter/en  of  lower  denominations. 

II.  Write  the  numbers  so  that  those  of  the  same  unit  value  will 
stand  in  the  same  column. 

III.  Beginning  at  the  right  hand,  add  each  denomination  as  in 


208  COMPOUND  NUMBERS. 

simple  numbers,  carrying  to  each  succeeding  denomination  one  for 
as  many  units  as  it  takes  of  the  denomination  added}  to  make  one 
of  the  next  higher  denomination. 

NOTE.  —  The  pupil  cannot  fait  to  see  that  the  principles  involved  in  adding 
compound  numbers  are  the  same  as  those  in  addition  of  simple  numbers;  and 
that  the  only  difference  consists  in  the  different  carrying  units. 

EXAMPLES    FOR   PRACTICE. 

(!•)  (2.) 

lb.   oz.  pwt.   gr.  H>.    5   5   3   gr. 

14  6  12  13         10   8  5  1   8 
17   5   3  12          7   7  6  2  13 

15  9  16          5  11  7 

2   7  15  20         21  10       16 

13   2   1  19          12   1  2  2   3 

4   1   5  21  7  1  19 

Sum,  66  11   9   5         58   4  5  2  19 

(3.)  (4.) 

mi.   rd.   ft.    in.  A.    P.  sq.  yd.  sq.  ft. 

7  26  11   9  140  137  27  6 

4  16   7  11  320   70  14  2 
36  14   3  111    7   3 

1928  214  95  22  7 

5  10       1  100  G  1 
625  25  76  8 
1     15     13 10  104  89  1  4 

5.  Add  1  T.  17  cwt.  8  lb.,  5  cwt.  29  lb.  8  oz.,  1  cwt.  42  lb. 
6  oz.,  and  17  lb.  8  oz.  Ans.  2  T.  3  cwt.  97  lb.  6  oz. 

6.  Add  6  yd.  2  ft.,  3  yd.  1  ft.  8  in.,  1  ft.  10*  in.,  2  yd.  2  ft. 
6*  in.,  2  ft.  7  in.,  and  2  yd.  5  in.  Ans.  16  yd.  2  ft.  1  in. 

7.  Add  4  Cd.  7  cd.  ft.,  2  Cd.  2  cd.  ft.  12  cu.  ft.,  6  cd.  ft.  15 
cu.  ft.,  5  Cd.  3  cd.  ft.  8  cu.  ft.,  and  2  Cd.  1  cu*  ft. 

8.  What  is  the  sum  of  If  hhd.  42  gal.  3  qt.  1}  pt.,  I  gal. 
2  qt.  f  pt.,  and  1.75  pt.  ?  Ans.  2  hhd.  23  gal.  2  qt.  3  gi. 

9.  What  is  the  sum  of  145$  A.,  7  A.,  109J  P.,  1  A.,  136.5  P., 
and  |  A.?  Ans.  156  A.  39£  P. 

10.  Required  the  sum  of  31  bu.  2  pk.,  10-J  bu.,  5  bu.  6£  qt., 
14  bu.  2.75  pk.,  and  f  pk.  Ans.  62  bu.  1  pk.  5  qt  If  pt. 


SUBTRACTION.  209 

11.  Required  the  value  of  42  yr.  7 A  mo.  -f  10  yr.  3  wk.  5  da. 
•f  9|  mo.  -J-  1  wk.  16  h.  40  min.  -f  |  mo.  +  3|  da. 

Am.  53  yr.  7  mo.  9  da.  23  h.  52  min. 

12.  Add  3  S.  22°  50',  24°  36'  25.7",  17'  18.2",  1  S.  3°  12' 
15.5",  12°  36'  17.8",  and  57.3".  Ans.  6  S.  3°  33'  14.5". 

13.  How  many  units  in  1^  gross  7^  doz.,  3  gross  1|  doz.,  |  of 
a  great  gross,  6|  doz.,  and  4  doz.  7  units  ?        •         Ans.  2183. 

14.  What  is  the  sum  of  240  A.  6  sq.  ch.,  212.1875  sq.  ch., 
and  5  sq.  ch.  lOf  P.?  Am.  262  A.  3  sq.  ch.  13.8  P. 

15.  Add  31  Pch.  18  cu.  ft.,  84.6  cu.  ft,,  |  Pch.,  and  f  °  cu.  ft. 

16.  Add  $3f ,  $25^,  $12J,  $2f ,  and  $2.54  j.   Ans.  $47.0725. 

17.  What  is  the  sum  of  3  ft  5  g  4  %  2  9  17  gr.,  2  It)  5  5  12 
gr.,43  2^19  16  gr.?  Ans.  5ft  lOg  4^29  5  gr. 

18.  A  N.  Y.  farmer  received  $.60  a  bushel  for  4  loads  of  corn ; 
the  first  contained  42.4  bu.,  the  second  2866  lb.,  the  third  36| 
bu.,  and  the  fourth  39  bu.  29  lb.     How  much  did  he  receive  for 
the  whole?  Ans.  $100.84-. 

19.  Bought  three  loads  of  hay  at  $8  per  ton.    The  first  weighed 
1.125  T.,  the  second  If  T.,  and  the  third  2500  pounds;  how  much 
did  the  whole  cost  ?  Ans.  $30.20. 

20.  A  man  in  digging  a  cellar  removed  140 J  cu.  yd.  of  earth, 
in  digging  a  cistern  24.875  cu.  yd.,  and  in  digging  a  drain  46  cu. 
yd.  20|  cu.  ft.     What  was  the  amount  of  earth  removed,  and  how 
much  the  cost  at  18  cts.  a  cu.  yd.  ? 

Ans.  212.425  cu.  yd.  removed;  $38.24  —  cost. 


SUBTRACTION. 

378.    1.  From  18  lb.  5  oz.  4  pwt.  14  gr.  take  10  lb.  6  oz.  10 
pwt.  8  gr. 

OPERATION.  ANALYSIS.      Writing    the    subtra- 

ib.         oz.        pwt.  gr.  hend  under   the   minuend,    placing 

5          4  14  units  of  the  same  denomination  under 

10          6        10  8  each  other,  we  subtract  8  gr.  from 

7        10        14  6  14  gr.  and  write  the  remainder,  6 

gr.,  underneath.     Since  we   cannot 

18*  O 


210  COMPOUND  NUMBERS. 

i 

subtract  10  pwt.  from  4  pwt.,  we  add  1  oz.  or  20  pwt.  to  the  4  pwt., 
subtract  10  pwt.  from  the  sum,  and  write  the  remainder,  14  pwt., 
underneath.  Having  added  20  pwt.  or  1  oz.  to  the  C  oz.  in  the  sub- 
trahend, we  find  that  we  cannot  subtract  the  sum,  7  oz.,  from  the  5 
oz.  in  the  minuend  ;  we  therefore  add  1  Ib.  or  12  oz.  to  the  5  oz.,  sub- 
tract 7  oz.  from  the  sum,  and  write  the  remainder,  10  oz.,  underneath. 
Adding  12  oz.  or  1  Ib.  to  the  10  Ib.  in  the  subtrahend,  we  subtract 
the  sum,  11  Ib.,  from  the  18  Ib.  in  the  minuend,  as  in  simple  numbers, 
and  write  the  remainder,  7  Ib.,  underneath. 

2.  From  12  bar.  15  gal.  3  qt.  take  7  bar.  18  gal.  1  qt. 

OPERATION.  ANALYSIS.     Proceeding  as  in  the  last 

bar.         gal.         qt.          operation,  we  obtain  a  remainder  of  4  bar. 
i;? 


28J  gal.  2  qt.     But,  £  gal.  =  2  qt.,  which 


added  to  the  2  qt.  in  the  remainder  makes 
28£  1  gal.,  and  this  added  to  the  28  gal.  makes 


4          29  29  gal.  ;  and  the  answer  is  4  bar.  29  gal. 

3.  From  f  of  a  rod  subtract  f  of  a  yard. 

OPERATION.  ANALYSIS.   We  first 

,     -,                ,    A  «.     .  ,  .  find  the  value  of  each 
|  rd.  =4  yd.  0  ft.  41  in. 

1  yd    =             2  "   3      «  fraction  in  integers  of 

4—~.  —  !  --  H  -  lower   denominations, 

1        lj  (369),  and  then  sub- 

Or,  tract    the    less   value 

•  from  the  greater-    Or> 


3  Ya   x        _  a  vd    v    2   -  -3   rd 

*  JQ      <    5£  -   4   )a     *    TT  -   *2  m- 

|  rd.  —  -^  rd.  =  |J  rd.  ;  given  fractions  to  frac- 

||  rci.  =  3  yd.  1  ft.  1J  in.  tions  °f  tne  same  de~ 

nomination,     subtract 

the  less  value  from  the  greater,  and  find  the  value  of  the  remainder 
in  integers  of  lower  denominations. 

379.  From  these  illustrations  we  deduce  the  following 
RULE.     I.  If  any  of  tJie  numbers  are  denominate  fractions,  or 
if  any  of  the  denominations  are  mixed  numbers,  reduce  the  frac- 
tions to  integers  of  lower  denominations. 

II.  Write  the  subtrahend  under  the  minuend,  so  that  units  of  the 
same  denomination  shall  stand  under  each  other. 

III.  Beginning  at  the  right  hand,  subtract  each  denomination 
separately,  as  in  simple  numbers. 


SUBTRACTION.  211 

IV.  //  the  number  of  any  denomination  in  the  subtrahend  ex- 
ceed that  of  the  same  denomination  in  the  minuend,  add  to  the 
number  in  the  minuend  as  many  units  as  make  one  of  the  next 
higher  denomination,  and  then  subtract  /  in  this  case  add  1  to  the 
next  higher  denomination  of  the  subtrahend  before  subtracting. 
Proceed  in  the  same  manner  with  each  denomination. 

EXAMPLES   FOR  PRACTICE. 


00 

(20 

mi. 

rd. 

ft. 

in. 

A. 

P. 

From  175 

147 

11 

4 

320 

146.4 

Take   59 

250 

12 

9 

150 

111.86 

Rem.  115 

216 

15 

1 

170 

34.54 

(3.) 

(4-j 

hhd.   gal. 

qt. 

yr. 

mo.  wk.   da. 

h. 

5   36 

H 

45 

1    3    0 

17£ 

2   45 

y 

10 

9   1   22 

6.8 

5.  Subtract  15  rd.  10  ft.  3J  in.  from  26  rd.  11  ft.  3  in. 

Ans.    11  rd.  llf  in. 

6.  From  1  T.  11  cwt.  30  Ibs.  6  oz.  take  18  cwt.  45  Ib. 

7.  Subtract  .659  wk.  from  2  wk.  3-jj-  da. 

Ans.  1  wk.  6  da.  5  h.  17  min.  16£  sec. 

8.  From  |fj  hhd.  take  .90625  gal.  Ans.  32  gal. 

9.  From  |  of  3|  A.  take  142.56  P. 

10.  Subtract  -fa  Ib.  Troy,  from  10  Ib.  8  oz.  8  pwt. 

11.  From  a  pile  of  wood  containing  36  Cd.  4  cd.  ft,  there  was 
•old  10  Cd.  6  cd.  ft,  12  cu.  ft;  how  much  remained? 

12.  From  5^-  barrels  take  %  of  a  hogshead. 

Ans.   4  bbl.  11  gal.  1  qt. 

1 3.  Subtract  -J-JJ  of  a  day  from  f  of  a  week. 

Ans.  4  da.  49  min.  30  sec. 

14.  From  -|  of  a  gross  subtract  %  of  a  dozen.      Ans.  6|-  doz. 

15.  From  %  of  a  mile  take  fj  of  a  rod. 

16.  Subtract  2  A.  125.76  P.  from  5  A.  64.24  P. 

Ans.  2  A.  98.48  P. 


212  COMPOUND  NUMBERS. 

17.  Subtract  .0625  bu.  from  |  pk.  'Ans.  4  qt. 

18.  From  the  sum  of  «  of  3651  da.  and  |  of  5§  wk.  take  49} 
min-  Ans.  33  wk.  1  da.  1  h.  10f  min. 

19.  From  the  sum  off  of  3|  mi.  and  17^  rd.,  take  21 3-J-  rd. 

20.  From  15  bbl.  3.25  gal.  take  14  bbl.  24  gal.  3.54  qt. 

21.  A  farmer  in  Ohio  having  200  bu.  of  barley,  sold  3  loads, 
the  first  weighing  1457  lb.,  the  second  1578  lb.,  and  the  third 
1420  lb. ;  how  many  bushels  had  he  left?     Ans.  107  bu.  9  lb. 

22.  Of  a  farm  containing  200  acres  two  lots  were  reserved,  one 
containing  50  A.  136.4  P.  and  the  other  48  A.  123.3  P.;  the  re- 
mainder was  sold  at  $35  per  acre.     How  much  did  it  brin<r  ? 

Ans.  $3513.19  + 

23.  An  excavation  58  ft.  long,  37  ft.  wide,  and  6  ft.  deep  is  to 
be  made  for  a  cellar;  after  471  cu. yd.  16  cu.  ft.   972  cu.  in.  of 
earth  have  been  removed,  how  much  more  still  remains  to  be 
taken  out?  Ans.     5  cu.  yd.  7  cu.  ft.  756  cu.  in. 

24.  From  the  sum  of  f  lb.,  4|  oz.,  and  311  pwt.,  take  the  differ* 
ence  between  f  oz.  and  |  pwt.     Ans.   1  lb.  3  oz.  8  pwt.  21  gr. 

2&.  From  the  sum  of  5T\-  A.,  f  of  6J  A.,  30  P.,  and  -fa  of  2^- 
P.,  take  4  A.  25  P.  12  sq.  yd. 

Ans.  5  A.  125  P.  6  sq.  yd. 

88O.   To  find  the  difference  in  dates. 

1.  How  many  years,  months,  days  and  hours  from  3  o'clock  p. 
M.  of  June  15,  1852,  to  10  o'clock  A.  M.  of  Feb.  22,  1860? 

OPERATION.  ANALYSIS.     Since  the  later  of  two  dates 

yr.       mo.    da.      h.  always   expresses  the  greater  period  of 

time,  we  write  the  later  date  for  a  minu- 
end and  the  earlier  date  for  a  subtrahend, 

78        6      19  placing  the  denominations  in  the  order  of 

the  descending  scale  from  left  to  right, 

(300,  NOTE  8).     We  then  subtract  by  the  rule  for  subtraction  of 
compound  numbers. 

When  the  exact  number  of  days  is  required  for  any  period  not 
exceeding  one  ordinary  year,  it  may  be  readily  found  by  the  fol- 
lowing 


SUBTRACTION. 


213 


TABLE, 

Showing  the  number  of  days  from  any  day  of  one  month  to  the  same 
day  of  any  other  month  within  one  year. 


FROM  ANT 

DAY  OF 

TO  THE  SAME  DAT  OF  THE  NEXT 

Jan. 

Feb. 

Mar. 

Apr. 

May 

June 

July. 

Aug. 

Sept. 

Oct. 

Nov. 

Dec. 

January   

365 
334 
306 
275 
245 
214 
184 
153 
122 
92 
61 
31 

31 
365 
337 
306 
276 
245 
215 
184 
153 
123 
92 
62 

59 
28 
365 
335 
304 
273 
243 
212 
181 
151 
120 
90 

90 
59 
31 
365 
335 
304 
274 
243 
212 
182 
151 
121 

120 
89 
61 
30 
365 
334 
304 
273 
242 
212 
181 
151 

151 

120 
92 
61 
31 
365 
335 
304 
273 
243 
212 
182 

181 
150 
122 
91 
61 
30 
365 
334 
303 
273 
242 
212 

212 
181 
153 
122 
92 
61 
31 
365 
334 
304 
273 
243 

243 
212 
184 
153 
123 
92 
62 
31 
365 
335 
304 
274 

273 
242 
214 
183 
153 
122 
92 
61 
30 
365 
334 
304 

304 
273 
245 
214 
184 
153 
123 
92 
61 
31 
365 
335 

334 
303 
275 
244 
214 
183 
153 
122 
91 
61 
30 
365 

February 

April  

May 

June  

July  

September  
October. 

November  

If  the  days  of  the  different  months  are  not  the  same,  the  num- 
ber of  days  of  difference  should  be  added  when  the  earlier  day 
belongs  to  the  month  from  which  we  reckon,  and  subtracted  when 
it  belongs  to  the  month  to  which  we  find  the  time.  If  the  29th 
of  February  is  to  be  included  in  the  time  computed,  one  day 
must  be  added  to  the  result. 


EXAMPLES    FOR   PRACTICE. 

1.  War  between  England  and  America  was  commenced  April 
19,  1775,  and  peace  was  restored  January  20,  1783;  how  long  did 
the  war  continue  ?  Ans.  7  yr.  9  mo.  1  da. 

2.  The  Pilgrims  landed  at  Plymouth  Dec.  22,  1620,  and  Gen. 
Washington  was  born  Feb.  22,  1782;  what  was  the  difference  in 
time  between  these  events  ? 

3.  The  first  settlement  made  in  the  U.  S.  was  at  Jamestown,  Va., 
May  23,  1607;  how  many  years  from  that  time  to  July  4,  1860  ? 

4.  How  long  has  a  note  to  run,  dated  Jan.  30,  1859,  and  made 
payable  June  3,  1861  ?  Ans.  2  yr.  4  mo.  3  da. 


214  COMPOUND   NUMBERS. 

t 

5.  How  many  years,  months,  and  days  from  your  birthday  to 

this  date  ? 

6.  Wh-at  length  of  time  elapsed  from  16  minutes  past  10  o'clock, 
A.  M  ,  July  4,  1855,  to  22  minutes  before  8  o'clock,  p.  M.,  Dec. 
12,  1860  ?  Am.  1988  da.  9  h.  22  min. 

7.  What  length  of  time  will  elapse  from  40  minutes  25  seconds 
past  12  o'clock,  noon,  April  21,  1860,  to  4  minutes  36  seconds 
before  5  o'clock,  A.  M.,  Jan.  1,  1862? 

8.  How  many  days  from  the  4th  September,  to  the  27th  of 
May  following  ?  Ans.  265  da. 


MULTIPLICATION. 

381.   1.  Multiply  5  mi.  178  rd.  15  ft.  by  6. 

ANALYSIS.    Writing  the  multi- 

OPE  RATION.  plier  under  the  lowest  denomi- 

5  mi.  178  rd.    15  ft,  nation  of  the  multiplicand,  we 

__  "  multiply  each  denomination   in 

33  113  7£  the    multiplicand    separately  in 

order  from  lowest  to  highest,  as 

in  simple  numbers,  and  carry  from  loAver  denominations  to  higher, 
according  to  the  ascending  scale  of  the  multiplicand,  as  in  addition 
of  compound  numbers.  Hence, 

RULE.     I.    Write  the  multiplier  under  the  loivest  denomination 
of  the  multiplicand. 

II.    Multiply  as  in  simple  numbers,  and  carry  as  in  addition  of 
compound  numbers. 

NOTES.  —  1.  When  the  multiplier  is  large,  nnd  is  a  composite  number,  we  may 
shorten  the  work  by  multiplying  by  the  component  factors. 

2.  The  multiplier  must  be  an  abstract  number. 

3.  If  any  of  the  denominations  are  mixed  numbers,  they  may  either  be  re- 
duced to  integers  of  lower  denominations  before  multiplying,  or  they  may  be 
multiplied  MS  directed  in  193. 

4.  The  multiplication  of  >i  denominate  fraction  is  the  most  readily  performed 
by  193,  :if'r,<.-r  which  the  product  may  bo  reduced  to  integers  of  lower  denomina- 
tions by  389. 


As  the  work  of  multiplying  by  large  prime  numbers  is 
somewhat  tedious,  the  following  method  may  often  be  so  modified 
and  adapted  as  to  greatly  shorten  the  operation. 


MULTIPLICATION. 


215 


1.  How  many  bushels  of  grain  in  47  bags,  each  containing  2  bu. 
1  pk.  4  qt.  ? 


FIRST    OPERATION. 

47  =  (5  x  9)  +  2 
2  bu.  1  pk.  4  qt.  x  2 
o 


1 1  bu.  o  pk.  4  qt.  in    5  bags. 
9 


106  bu.  3  pk.  4  qt.  in  45  bags. 
4   «    3    *<  "2     " 

111  bu.  2  pk.~4  qt.   "  47     " 


SECOND    OPERATION. 

47  =  (6  x  8)  _  1 
2  bu.  1  pk.  4  qt,  x  1 
6 


14  bu.  1  pk.  in    G  bags. 


114  bu.  in  48  bags. 

2    "    1  pk.  4  qt.  "     1  bag. 

Ill  bu.  2  pk.  4  qt.  "  47  bags. 


ANALYSIS.  Multiplying 
the  contents  of  1  bag  by  5, 
and  the  resulting  product  by 
9,  we  have  the  contents  cf 
45  bags,  which  is  the  com- 
posite number  next  less  than 
the  given  prime  number,  47. 
Next,  multiplying  the  con- 
tents of  1  bag  by  2,  we  have 
the  contents  of  2  bags,  -which, 
added  to  the  contents  of  45 
bags,  gives  us  the  contents 
of  45  -f  2  =  47  bags. 

Or,  we  may  multiply  the 
contents  of  1  bag  by  the  fac- 
tors of  the  composite  num- 
ber next  greater  than  the 
given  prime  number,  47,  and 
from  the  last  product  sub- 
tract the  multiplicand. 


EXAMPLES    FOR   PRACTICE. 


(1.) 

T.       cwt.        Ib.       oz. 

12        15        27        9 

8 


102 

2       20 

8 

(3.) 

A. 

P.       Bq.  yd. 

Bq.  ft. 

7 

73         21 

7 

6 

mi. 
14 


(2.) 

rd.  ft. 

276        14 

9 


133       251 


Cd.     cd.  ft.    cu.  ft. 
10        7        13 
12 


5.  Multiply  34  bu.  3  pk.  6  qt.  1  pt.  by  14. 

6.  Multiply  4  Ib.  10  oz.  18.7  pwt,  by  27. 

Ans.  132  Ib.  7  oz.  4.9  pwt 

7.  Multiply  95  33  2  3  13  gr.  by  35. 


216  COMPOUND    NUMBERS. 

8.  Multiply  5  gal.  2  qt,  1  pt.  3.25  gi.  by  96. 

9.  Multiply  78  A.  135  P.  15  sq.  yd.  by  15-f. 

Ans.  1235  A.  42  P.  23}  sq.  yd. 

10.  What  is  73  times  9  cu.  yd.  10  cu.  ft.  1424  cu.  in.? 

11.  Multiply  2  Ib.  8  oz.  13  pwt.  18  gr.  by  59. 

12.  Multiply  4  yd.  1  ft.  4.7  in.  by  125. 

13.  If  1  qt.  2  gi.  of  wine  fill  1  bottle,  how  much  will  be  re- 
quired to  fill  a  gross  of  bottles  of  the  same  capacity  ? 

14.  Multiply  7  0.  10  f  I  4  f  3  25  1t[  by  24. 

Ans.  22  Cong.  7  0.  13  f  g  2  f  3. 

15.  Multiply  3  hhd.  43  gal.  2.6  gi.  by  17. 

16.  Multiply  9  T.  13  cwt.  1  qr.  10.5  Ib.  by  1.7. 

NOTE. — When  the  multiplier  contains  a  decimal,  the  multiplicand  may  be  re- 
duced to  the  lowest  denomination  mentioned,  or  the  lower  denominations  to  a 
decimal  of  the  higher,  before  multiplying.  The  result  can  be  reduced  to  the 
compound  number  required. 

Ans.  16T.  8  cwt.  2qr.  20.65  Ib. 

17.  If  a  pipe  discharge  2  hhd.  23  gal.  2  qt.  1  pt.  of  water  in  1 
hour,  how  much  will  it  discharge  in  4.8  hours,  if  the  water  flow 
with  the  same  velocity?          Ans.  11  hhd.  25  gal.  1  pt.  2.4  gi. 

18.  What  will  be   the  value  of  1  dozen  gold  cups,  each  cup 
weighing  9  oz.  13  pwt.  8  gr.,  at  $212.38  a  pound  ? 

19.  What  cost  5  casks  of  wine,  each  containing  27  gal.  3  qt.  1 
pt.,  at  $1.371  a  gallon?  Ans.  $191.64  + . 

20.  A  farmer  sold  5  loads  of  oats,  averaging  37  bu.  3  pk.  5  qt. 
each,  at  $.65  per  bushel;  how  much  money  did  he  receive  for  the 
grain?  Ans.  $123.20—. 

DIVISION. 
383.   1.  Divide  37  A.  60  P.  7  sq.  yd.  by  8. 

ANALYSIS.    Wri  ting  the  divisor  on 

OPERATION,       j  the  left  of  the  Dividend,  divide  the 

8)37  A._60  P.  7  sq.  yd.          llighest  denomination,  and  the  quo- 
4      107      16  tient  is  4  A.  and  a  remainder  of  5  A. 

Write  the  quotient  under  the  de- 
nomination divided,  and  reduce  the  remainder  to  rods,  making  800  P., 
which  added  to  the  60  P.  of  the  dividend,  equals  860  P.  Dividing  this, 
we  have  a  quotient  of  107  P.  and  a  remainder  of  3  P.  Writing  the 
107  P.  under  the  denomination  divided,  we  reduce  the  remainder  to 


DIVISION. 

square  yards,  making  121  sq.  yd.,  which  added  to  the  7  sq.  yd.  of  the 
dividend,  equals  128  sq.  yd.  Dividing  this,  we  have  a  quotient  of  16 
sq.  yd.  and  no  remainder. 

2.  Divide  111  bu.  2  pk.  4  qt.  by  47. 

OPERATION. 

47)lllbu.2pk.4qt.(2bu.lpk.4qt  ANALYSIS.     The 

divisor  being  large, 

17  bu.  rem.  and  a  prime  num- 

4  her,  we  divide  by 

70  pk.  in  17  bu.  2  pk.  long  division,  set- 

47  ting  down  the 

whole  work  of  sub- 

2j>Pk-rem-  tractingandreduc- 

ing. 

188  qt.  in  23  pk.  4  qt 

188 

From  these  examples  and  illustrations  we  derive  the  following 
RULE.     I.   Divide  the  highest  denomination  as  in  simple  num- 
bers, and  each  succeeding  denomination  in  the  same  manner,  if 
there  be  no  remainder. 

II.  If  there  be  a  remainder  after  dividing  any  denomination, 
reduce  it  to  the  next  lower  denomination,  adding  in  the  given  num- 
ber of  that  denomination,  if  any,  and  divide  as  before. 

III.  The  several  partial  quotients  will  be  the  quotient  required. 

NOTES.  —  1.  When  the  divisor  is  large  and  is  a.  composite  number,  we  may 
shorten  the  work  by  dividing  by  the  factors. 

2.  When   the  divisor  and  dividend  are  both  compound  numbers,  they  must 
both  be  reduced  to  the  same  denomination  before  dividing,  and  then  the  process 
is  the  smne  as  in  simnle  numbers. 

3.  The  division  of  a  denominate  fraction  is  most  readily  performed  by  195, 
after  which  the  quotient  tnoy  bo  reduced  to  its  equivalent  compound  number, 
by  369. 

EXAMPLES   FOR   PRACTICE. 


£ 

5)62 

a-) 

8. 

7 

d. 

9 

far. 

3 

ib. 
9)56 

oz 

(20 

pwt. 

17 

gr. 

6 

Quotient,    12 
19 

9 

6 

8 

6Ib. 

3 

oz.     8  pw.  14  gr. 

218  COMPOUND  NUMBERS. 

(3.)  (4'.) 

hhd.      gal.       qt.        pt.  T.  cwt.        qr.         Ib. 

12)9      28      2  19)373       19 2 4_ 

"49      2       1  ~19       13       2      16 

5.  Divide  358  A.  57  P.  G  sq.  yd.  2  sq.  ft.  by  7. 

Ans.  51  A.  31  P.  8  sq.  ft. 

6.  Divide  192  bu.  3  pk.  1  qt.  1  pt.  by  9. 

7.  Divide  336  yd.  4  ft.  3^  in.  by  21.       Ans.  16  yd.  2|  in. 

8.  Divide  77  sq.  yd.  5  sq.  ft.  82  sq.  in.  by  13. 

Ans.  5  sq.  yd.  8  sq.  ft.  106  sq.  in. 

9.  Divide  678  cu.  yd.  1  cu.  ft.  1038.05  cu.  in.  by  67. 

10.  Divide  1986  mi.  3  fur.  20  rd.  1  yd.  by  108. 

11.  Divide  12  sq.  mi.  70  P.  by  22£. 

Ans.  341  A.  56f  P. 

NOTE  4. — Observe  that  22$  =  *•/  ;  hence,  multiply  by  2,  and  divide  the  result 
by  45. 

12.  Divide  365  da.  6  h.  by  240. 

13.  Divide  3794  cu.  yd.  20  cu.  ft.  709^  cu.  in.  by  33f 

14.  Divide  121  Ib.  3^  2^  19  4  gr.  by  13|. 

15.  Divide  28°  51'  27.765"  by  2.754.      Ans.  10°  28'  42*". 

16.  Divide  202  yd.  1  ft.  6|  in.  by  f . 

Ans.  337  yd.  1  ft,  7£  in. 

17.  Divide  1950  da.  15  h.  15|  min.  by  100. 

18.  If  a  town  4  miles  square  be  divided  equally  into  124  farms, 
how  much  will  each  farm  contain  ?  Ans.  82  A.  92ff  P. 

19.  A  cellar  48  ft.  6  in.  long,  24  ft.  wide,  and  6*  ft.  deep,  was 
excavated  by  6  men  in  8  days;  how  many  cubic  yards  did  each 
man  excavate  daily?          Ans.  5  cu.  yd.  22  cu.  ft.  1080  cu.  in. 

20.  How  many  casks,  each  containing  2  bu.  3  pk.  6  qt.,  can  be 
filled  from  356  bu.  3  pk.  5.qt.  of  cherries?  Ans.  12 


LONGITUDE   AND  TIME. 

SJ84L  Since  the  earth  performs  one  complete  revolution  on  its 
axis  in  a  day  or  24  hours,  the  sun  appears  to  pass  from  east  to 
west  round  the  earth,  or  through  360°  of  longitude  once  in  every 


LONGITUDE  AND  TIME.  219 

24  hours  of  time.     Hence  the  relation  of  time  to  the  real  motion 
of  the  earth  or  the  apparent  motion  of  the  sun,  is  as  follows: 

Time.  Longitude. 

24  h.  =     360° 

1  h.  or  60  min.  =    8^°°      =     15°    =    900/ 

1  min.  or  60  sec.    =   ^°  =    %(/       =     15'    =    900" 

Isec.                       =   if  =    V*°"     =     l?x/ 

Hence,  1  h.  of  time  -  =    15°  of  longitude. 

1  min.     "  =    15X  "           " 

1  sec.      "  =    15" "           " 

CASE    I. 

385.  To  find  the  difference  of  longitude  between  twc 
places  or  meridians,  when  the  difference  of  time  is 
known. 

ANALYSIS.  A  difference  of  1  h.  of  time  corresponds  to  a  difference 
of  15°  of  longitude,  of  1  min.  of  time  to  a  difference  of  15'  min.  of 
longitude,  and  of  1  sec.  of  time  to  a  difference  of  15"  of  longitude, 
(384).  Hence,  the 

RULE.  Multiply  the  difference  of  time,  expressed  in  hours, 
minutes,  and  seconds,  l>y  15,  according  to  the  rule  for  multiplica- 
tion of  compound  numbers  ;  the  product  will  be  the  difference  of 
longitude  in  degrees,  minutes,  and  seconds. 

NOTES.  —  1.  If  one  place  be  in  east,  and  the  other  in  west  longitude,  the  dif- 
ference of  longitude  is  found  by  adding  them,  and  if  the  sum  be  greater  than 
180°,  it  must  be  subtracted  from  360°. 

2.  Since  the  sun  appears  to  move  from  east  to  west,  when  it  is  exactly  12 
o'clock  sit  one  place,  it  will  be  past  12  o'clock  at  all  places  east,  and  before  12  ;.t 
nil  places  west.  Hence,  if  the  difference  of  time  between  two  places,  be  subtracted 
from  the  time  at  the  eagerly  place,  the  result  will  be  the  time  at  the  westerly 
place;  and  if  the  difference  be  added  to  the  time  at  the  westerly  place  the  result 
will  be  the  time  at  the  easterly  place. 

EXAMPLES    FOR    PRACTICE. 

1.  When  it  is  9  o'clock  at  Washington,  it  is  8  h.  7  min.  4  sec. 
at  St.  Louis;  the  longitude   of  Washington  being  77°  1',  west, 
•what  is  the  longitude  of  St.  Louis?  Ans.  90°  15'  west. 

2.  The  sun  rises  at  Boston   1  h.   11  min.  56  sec.  sooner  than 
at  New  Orleans;  the  longitude  of  New  Orleans  being  89°  2'  west, 
what  is  the  longitude  of  Boston  ?  Ans.  71°  3'  west. 


220  COMPOUND  NUMBERS. 

i 

3.  When  it  is  half  past  2  o'clock  in  the  morning  at  Havana,  it 

is  9  h.  13  min.  20  sec.  A.  M.  at  the  Cape  of  Good  Hope;  the 
longitude  of  the  latter  place  being  18°  28'  east,  what  is  the 
longitude  of  Havana  ?  Am.  82°  22'  west. 

4.  The  difference  of  time  between  Valparaiso  and  Rome  is  6  h. 
8  min.  28  sec. ;  what  is  the  difference  in  longitude  ? 

5.  A  gentleman  traveling  East  from  Fort  Leavenworth,  which 
is  in  94°  44'  west  longitude,  found,  on  arriving  at  Philadelphia,  that 
his  watch,  an  accurate  time  keeper,  was  1  h.  18  min.  16  sec.  slower 
than  the  time  at  Philadelphia;  what  is  the  longitude  of  Philadel- 
phia ?  Ans.  75°  10'  west. 

6.  When  it  is  12  o'clock  M.  at  San  Francisco  it  is  2  h.  58  min. 
23£  sec.  P.  M.  at  Rochester,  N,  Y;  the  longitude  of  the  latter 
place  being  77°  51'  W.,  what  is  the  longitude  of  San  Francisco? 

7.  A  gentleman  traveling  West  from  Quebec,  which  is  in. 71° 
12'  15"  W.  longitude,  finds,  on  his  arrival  at  St.  Joseph,  that  his 
watch  is  2  h.  33  min.  53{  J  sec.  faster  than  true  time  at  the  latter 
place.     If  his  watch  has  kept  accurate  time,  what  is  the  longitude 
of  St.  Joseph  ?  Ans.  109°  40'  44"  V. 

8.  A  ship's  chronometer,  set  at  Greenwich,  points  to  5  h.  40 
min.  20  sec.  p.  M.,  when  the  sun  is  on  the  meridian ;  what  is  the 
ship's  longitude  ?  Ans.  85°  5'  west. 

NOTK  3. — Greenwich,  Eng.,  is  on  the  meridian  of  0°,  and  from  this  meridian 
longitude  is  reckoned. 

9.  The  longitude  of  Stockholm  being  18°  3'  30"  E.,  when  it  is 
midnight  there,  it  is  5  h.  51  min  41  ?  sec.  p.  M.  at  New  York;  what 
is  the  longitude  of  New  York  from  Greenwich  ? 

An*.  74°  r  <>"  W 

10.  A  vessel  set  sail  from  New  York,  and  proceeded  in  a  south- 
easterly direction  for  24  days.     The  captain  then  took  an  obser- 
vation on  the  sun,  and  found  the  local  time  at  the  ship's  meridian 
to  be  10  h.  4  min.  36.8  sec.  A.  M. ;  at  the  moment  of  the  observa- 
tion, his  chronometer,  which  had  been  set  for  New  York  time, 
showed  8  h.  53  min.  47  sec.  A.  M.    Allowing  that  the  chronometer 
had  gained  3.56  sec.  per  day,  how  much  had  the  ship  changed  her 
longitude  since  she  set  sail?  Ans.    18°  3'  48.6". 


LONGITUDE  AND  TIME.  221 


CASE   II. 

386.  To  find  the  difference  of  time  between  two 
places  or  meridians,  when  the  difference  of  longitude 
is  known. 

ANALYSIS.  A  difference  of  15°  of  longitude  produces  a  difference 
of  1  h.  of  time,  15'  of  longitude  a  difference  of  1  min.  of  time,  and 
15"  of  longitude  a  difference  of  1  sec.  of  time,  (384).  Hence  the 

RULE.  Divide  the  difference  of  longitude,  expressed  in  degrees, 
minutes,  and  seconds,  l>y  15,  according  to  the  rule  for  division  of 
compound  numbers;  the  quotient  will  be  the  difference  of  time  in 
hours,  minutes,  and  seconds. 

EXAMPLES    FOR   PRACTICE. 

1.  Washington  is  77°  V  and  Cincinnati  is  84°  24'  west  longi- 
tude; what  is  the  difference  of  time?         Ans.  29  min.  32  sec. 

2.  Paris  is  2°  20'  and  Canton  113°  14'  east  longitude;  what  is 
the  difference  in  time  ? 

3.  Buffalo  is  78°  55'  west,  and  the  city  of  Rome  20°  30'  east 
longitude ;  what  is  the  difference  in  time  ? 

Ans.  6  h.  37  min.  40  sec. 

4.  A  steamer  arrives  at  Halifax,  63°  36'  west,  at  4  h.  30  min. 
p.  M.  ;  the  fact  is  telegraphed  to  New  York,  74°  V  west,  without 
loss  of  time ;  what  is  the  time  of  its  receipt  at  New  York  ? 

5.  The  longitude  of  Cambridge,  Mass.,  is  71°  7'  west,  and  of 
Cambridge,  England,  is  5'  2"  east;  what  time  is  it  at  the  former 
place  when  it  is  12  M.  at  the  latter  ? 

Ans.   7  h.  15  min.  llyf  sec.  A.  M. 

6.  The  longitude  of  Pekin  is  118°   east,  and  of  Sacramento 
City  120°  west;  what  is  the  difference  in  time? 

7.  The  longitude  of  Jerusalem  is  35°  32'  east,  and  that  of 
Baltimore  76°  37'  west;  when  it  is  40  minutes  past  6  o'clock 
A.  M.  at  Baltimore,  what  is  the  time  at  Jerusalem? 

8.  What  time  is  it  in  Baltimore  when  it  is  6  o'clock  p.  M.  at 
Jerusalem?  Ans.  10  h.  31  min.  24  sec.  A.  M. 

19* 


222  COMPOUND  NUMBERS. 

9.  The  longitude  of  Springfield,  Mass.,  is  72°  35'  45"  W.,  and 
of  Galveston,  Texas,  94°  46'  34"  W.;  when  it  is  20  inin.  past  6 
o'clock  A.  M.  at  Springfield,  what  time  is  it  at  Galveston  ? 

10.  The  longitude  of  Constantinople  is  28°  49'  east,  and  of  St. 
Paul  93°  5'  west;  when  it  is  3  o'clock  p.  M.  at  the  latter  place, 
what  time  is  it  at  the  former  ? 

11.  What  time  is  it  at  St.  Paul  when  it  is  midnight  at  Constan- 
tinople? Ans.  3  h.  52  miri.  24  sec.  r.  M. 

12.  The  longitude  of  Cambridge,  Eng.,  is  5'  2"  E.,  and  of 
Mobile,  Ala.,  88°  1'  29"  W.;  when  it  is  12  o'clock  M.  at  Mobile, 
what  is  the  time  at  Cambridge  ? 

PROMISCUOUS    EXAMPLES    IN    COMPOUND   NUMBERS. 

1.  In  9  Ib.  8g  13  29  19  gr.  how  many  grains? 

2.  How  much  will  3  cwt.  12  Ib.  of  hay  cost,  at  $15^  a  ton? 

3.  In  27  yd.  2  qr.  how  many  Eng.  ells?  Ans.   22. 

4.  Reduce  818.945  to  sterling  money  Ans.  £3  17s.  10-|f|fd. 

5.  In  4  yr.  48  da.  10  h,  45  sec.  how  many  seconds  ? 

6.  How  many  printed  pages,  2  pages  to  each  leaf,  will  there  be 
in  an  octavo  book  having  24  fully  printed  sheets  ?      Ans.  384. 

7.  At  1/6  sterling  per  yard,  how  many  yards  of  cloth  may  be 
bought  for  £5  6s.  6d.  ?  Ans.  71  yd. 

8.  In  4  mi.  51  ch.  73  1.  how  many  links? 

9.  In  22  A.  153  sq.  rd.  2j-  sq.  yd.  how  many  square  yards? 

10.  How  many  demijohns,  each  containing  3  gal.  1  qt.  1  pt., 
can  be  filled  from  3  hhd.  of  currant  wine  ?  Ans.  56. 

11.  Paid  $375.75  for  2^  tons  of  cheese,  and  retailed  it  at  9£ 
cts.  a  pound ;  how  much  was  my  whole  gain  ? 

12.  A  gentleman  sent  a  silver  tray  and  pitcher,  weighing  3  Ib. 
9  oz.,  to  a  jeweler,  and  ordered  them  made  into  tea  spoons,  each 
weighing  1  oz.  5  pwt.  j  how  many  spoons  ought  he  to  receive  ? 

Ans.  3  doz. 

13.  What  part  of  4  gal.  3  qt.  is  2  qt,  1  pt.  2  gi.  ?       Ans.  4  J. 

14.  Reduce  |  of  T4T  of  a  rod  to  the  fraction  of  yard. 

15.  How  many  yards  of  carpeting  1  yd.  wide,  will  be  required 
to  cover  a  floor  26£  ft.  long,  and  20  ft.  wide  ?  Ans.  58|. 


PROMISCUOUS   EXAMPLES.  223 

16.  If  I  purchase  15  T.  3  cwt.  3  qr.  24  Ib.  of  English  iron,  by 
long  ton  weight,  at  6  cents  a  pound,  and  sell  the  same  at  8140  per 
short  ton,  how  much  will  I  gain  by  the  transaction  ? 

17.  What  will  be  the  expense  of  plastering  a  room  40  ft.  long, 
30 £  ft.  wide,  and  22  J-  ft.  high,  at  18  cents  a  sq.  yd.,  allowing  1375 
sq.  ft.  for  doors,  windows,  and  base  board?  Ans.  $69. 78£. 

18.  How  much  tea  in  23  chests,  each  weighing  78  Ib.  9  oz.  ? 

19.  Valparaiso  is  in  latitude  33°  2'  south,  and  Mobile  30°  41 
north ;  what  is  their  difference  of  latitude  ?  Ans.  63°  43'. 

20.  If  a  druggist  sell  1  gross  4  dozen  bottles  of  Congress  water 
a  day,  how  many  will  he  sell  during  the  month  of  July  ? 

21.  Eighteen  buildings  are  erected  on  an  acre  of  ground,  each 
occupying,  on  an  average,  4  sq.  rd.  120  sq.  ft.  84  sq.  in. ;  how 
much  ground  remains  unoccupied  ? 

22.  At  $13  per  ton,  how  much  hay  may  be  bought  for  $12.02 £  ? 

23.  If  1  pk.  4  qt.  of  wheat  cost  $.72,  how  much  will  a  bushel 
cost?  Ans.  $1.92. 

24.  How  many  bushels,  Indiana  standard,  in  36244  Ibs.  of 
wheat  ? 

25.  At  20  cents  a  cubic  yard,  how  much  will  it  cost  to  dig  a 
cellar  32  ft.  long,  24  ft.  wide,  and  6  ft.  deep?    Ans.  $34.13  +  . 

26.  If  the  wall  of  the  same  cellar  be  laid  l£  feet  thick,  what 
will  it  cost  at  $1.25  a  perch  ?  Ans.  |50.90jf . 

27.  The  forward  wheels  of  a  wagon  are  10   ft.  4  in.  in  circum- 
ference, and  the  hind  wheels  15 1  ft. ;  how  many  more  times  will 
the  forward  wheels  revolve  than  the  hind  wheels  in  running  from 
Boston  to  N.  Y.,  the  distance  being  248  miles?        Ans.  42240. 

28.  Bought  15  cwt.  22  Ib.  of  rice  at  $3.75  a  cwt,  and  7  cwt. 
36  Ib.  of  pearl  barley  at  $4.25  a  cwt.     How  much  would  be  gained 
by  selling  the  whole  at  4£  cents  a  pound?  Ans.  $13.255. 

29.  From  f  of  3  T.  10  cwt.  subtract  T4^  of  7  T.  3  cwt.  26  Ib. 

30.  What  is  the  value  in  avoirdupois  weight  of  16  ib.  5  oz.  10 
pwt.  13  gr.  Troy?  Ans.  13  Ib.  8  oz.  11.4-fdr. 

31.  What  decimal  of  a  rod  is  1  ft.  7.8  in.  ? 

32.  If  a  piece  of  timber  be  9  in.  wide  and  6  in.  thick,  what 
length  of  it  will  be  required  to  make  3  cu.  ft.  ?  Ans,  8  ft. 


224  COMPOUND  NUMBERS. 

33.  If  a  board  be  16  in.  broad,  what  length  of  it  will  make  7 
sq.  ft.  ?  Ans.  5i  ft. 

34.  If  a  hogshead  contain  10  cubic  feet,  how  many  more  gal- 
lons of  dry  measure  will  it  contain  than  of  beer  measure  ? 

35.  How  many  tons  in  a  stick  of  hewn  timber  60  ft.  long,  and  1 
ft.  9  in.  by  1  ft.  1  in.  ?  Ans.  2.275  tons. 

71 

36.  Subtract  ^  bu.  -f  |  of  f  f  of  3i  qt.  from  5  bu.  3f  1  qt. 

Ans.  16f  pk. 

37.  What  is  the  difference  between  -J  of  5  sq.  mi.  250  A.  120 
P.,  and  3£  times  456  A.  134  P.  25  sq.  yd.? 

Ans.  2  sq.  mi.  254  A.  106  P.  24-f  sq.  yd. 

38.  How  many  pounds  of  silver,  Troy  weight,  are  equivalent 
in  value  to  5.6  Ib.  of  gold  by  the  English  government  standard  ? 

Ans.  80  Ib.  2  pwt.  19.2768  gr. 

39.  If  a  piece  of  gold  is  $  pure,  how  many  carats  fine  is  it  ? 

40.  In  gold  16  carats  fine  what  part  is  pure,  and  what  part  is 
alloy  ? 

41.  A  man  having  a  piece  of  land  containing  384J  A.,  divided 
it  between  his  two  sons,  giving  to  the  elder  22  A.  1  II.  20  P.  more 
than  to  the  younger ;  how  many  acres  did  he  give  to  each  ? 

Ans.  203  A.  2  R.  14  P.,  elder ;  181  A.  0  R.  34  P.,  younger. 

42.  4000  bushels  of  corn  in  Illinois  is  equal  to  how  many  bushels 
in  New  York  ?  Ans.  3586/^  bu. 

43.  The  market  value  being  the  same  in  both  States,  a  farmer 
in    New  Jersey  exchanged  110    bu.  of  cloverseed,  worth  $4  a 
bushel,  with  a  farmer  in  New  York  for  corn,  worth  $f  a  bushel, 
which  he  sold  in  his  own  State  for  cash.     The  exchange  being 
made  by  weight,  in  whose  favor  was  the  difference,  and  how  much 
in  cash  value  ? 

Ans.  The  N.  J.  farmer  gained  69^  bu.  corn,  worth  $46^-. 

44.  The  great  pyramid  of  Cheops  measures  763.4  feet  on  each 
side  of  its  base,  which  is  square.    How  many  acres  does  it  cover  ? 

45.  The  roof  of  a  house  is  42  ft.  long,  and  each  side  20  ft.  6 
in.  wide;  what  will  the  roofing  cost  at  ^4.62^  a  square  ? 


PROMISCUOUS   EXAMPLES.  225 

46.  If  17  T.  15  cwt.  62£  Ib.  of  iron  cost  $1333.593,  how  much 
will  1  ton  cost? 

47.  How  many  wine  gallons  will  a  tank  hold,  that  is  4  ft.  long 
by  of  ft.  wide.,  and  If  ft.  deep?  Ans.  187^  gal. 

48.  What  will  be  the  cost  of  300  bushels  of  wheat  at  9s.  4d. 
per  bushel,  Michigan  currency  ?  Ans.  $350. 

49.  What  will  be  the  cost  in  Missouri  currency? 

50.  What  will  be  the  cost  in  Delaware  currency  ? 

61.   What  will  be  the  cost  in  Georgia  currency?    Ans.  $600. 

52.  What  will  be  the  cost  in  Canada  currency?     Ans.  $560. 

53.  Bought  the  following  bill  of  goods  in  Boston  : 

6£  yd.  Irish  linen      @  5/4 

12       «    flannel  "    3/9 

8i    "    calico  "    1/7 

9      "    ribbon  "     /9 

4*  Ib.  coffee  «    1/5 

6f  gal.  molasses  "    3/8 

What  was  the  amount  of  the  bill?  Ans.  $21.76  -f . 

54.  How  many  pipes   of  Madeira   are  equal  to  22  pipes   of 
sherry  ? 

55.  A  cubic  foot  of  distilled  water  weighs  1000  ounces  avoirdu- 
pois; what  is  the  weight  of  a  wine  gallon  ?     Ans.  8  Ib.  5^-J  oz. 

56.  There  is  a  house  45  feet  long,  and  each  of  the  two  sides 
of  the  roof  is  22  feet  wide.     Allowing  each  shingle  to  be  4  inches 
wide  and  15  inches  long,  and  to  lie  one  third  to  the  weather,  how 
many  half-thousand  bunches  will  be  required  to  cover  the  roof? 

Ans.  28-^V 

57.  A  cistern  measures  4  ft.  6  in.  square,  and  6  ft.  deep;  how 
many  hogsheads  of  water  will  it  hold  ? 

58.  If  the  driving  wheels  of  a  locomotive  be  18  ft.  9  in.  in  cir- 
cumference, and  make  3  revolutions  in  a  second,  how  long  will  the 
locomotive  be  in  running  150  miles? 

Ans.  3  h.  54  min.  40  sec. 

59    In  traveling,  when  I  arrived  at  Louisville  my  watch,  which 
was  exactly  right  at  the  beginning  of  my  journey,  and  a  correct 

P 


226  COMPOUND  NUMBERS. 

timekeeper,  was  1  h.  6  min.  52  sec.  fast;  from  .-what  direction 
had  I  come,  and  how  far  ?  Aits.  From  the  east,  16°  43'. 

60.  How  many  U.  S.  bushels  will  a  bin  contain  that  is  8.5  ft. 
long,  4.25  ft.  wide,  and  3f  ft.  deep? 

61.  Reduce  3  hhd.  9  gal.  3  qt.  wine  measure  to  Imperial  gal- 
lons. Am.  165.5807+  Imp' 1  gal. 

62.  A  man  owns  a  piece  of  land  which  is  105  ch.  85  1.  long, 
and  40  ch.  15  1.  wide;  how  many  acres  does  it  contain  ? 

63.  A  and  B  own  a  farm  together;  A  owns  T72  of  it  and  B  the 
remainder,  and  the  difference  between  their  shares  is  15  A.  68 \  P. 
How  much  is  B's  share?  Ans.  38  A.  91 J  P. 

64.  At  $3.40  per  square,  what  will  be  the  cost  of  tinning  both 
sides  of  a  roof  40  ft.  in  length,  and  whose  rafters  are  20  ft.  6  in. 
long?  Ans.  $55.76. 

65.  What  is  the  value  of  a  farm  189.5  rd.  long  and  150  rd. 
wide,  at  $3 If  per  acre? 

66.  Reduce  9.75  tons  of  hewn  timber  to  feet,  board  measure, 
that  is,  1  inch  thick.  Ans.   5850  ft. 

67.  How  many  wine  gallons  will  a  tank  contain   that  is  4  ft. 
long,  34  ft.  wide,  and  2f  ft.  deep  ?  Ans.  299^  gal. 

68.  If  a  load  of  wood  be  12  ft.  long,  and  3  ft.  6  in.  wide,  how 
high  must  it  be  to  make  a  cord  ? 

69.  In  a  school  room  32  ft.  long,  18  ft.  wide,  and  12  ft.  6  in. 
high,  are  60  pupils,  each  breathing  10  cu.  ft.  of  air  in  a  minute. 
In  how  long  a  time  will  they  breathe  as  much  air  as  the  room 
contains  ? 

70.  A  man  has  a  piece  of  land  201|  rods  long  and  4H  rods 
wide,  which  he  wishes  to  lay  out  into  square  lots  of  the  greatest 
possible  size.     How  many  lots  will  there  be  ?  Ans.  396. 

71.  A  man  has  4  pieces  of  land  containing  4  A.  140  P.,  6  A. 
132  P.,  9  A.  120  P.,  and  11  A.  112  P.  respectively.     It  is  re- 
quired to  divide  each  piece   into  the  largest  sized  building  lots 
possible,  each  lot  containing  the  same  area,  and  an  exact  number 
of  square  rods.     How  much  laud  will  each  lot  contain  ? 

Ans.  156  P. 


DUODECIMALS.  227 


DUODECIMALS. 

387.  Duodecimals  are  the  parts  of  a  unit  resulting  from  con- 
tinually dividing  by  12 ;  as  1,  rJ3,  yl^,  iVss*  e^c-     ^n  practice, 
duodecimals  are  applied  to  the  measurement  of  extension,  the  foot 
being  taken  as  the  unit. 

In  the  duodecimal  divisions  of  a  foot,  the  different  orders  of 
units  are  related  as  follows  : 

V    (inch  or  prime) is   ^   of  a  foot,  or  1  in.  linear  measure. 

I"  ( second )  or  -^  of  T'-2 , "  T£T  of  a  foot,  or  1  « '    square       ' ' 

V"  (third)  or  -^  of  ^/of  ^2-, . .  "  TJ-0^  of  a  foot,  or  1  "    cubic         " 

TABLE. 

12  fourths,  (""},  make  1  third I'" 

12  thirds  "      1  second, 1" 

12  seconds  "      1  prime, V 

12  primes,  "      1  foot, ft. 

SCALE  —  uniformly  12. 

The  marks  ',  ",  '",  "",  are  called  indices. 

NOTES.  —  1.  Duodecimals  are  really  common  fractions,  and  can  always  be 
treated  as  such ;  but  usually  their  denominators  are  not  expressed,  and  they  are 
treated  as  compound  numbers. 

2.  The  word  duodecimal  is  derived  from  the  Latin  term  duodecim,  signifying 
12. 

ADDITION  AND  SUBTRACTION. 

388.  Duodecimals  are  added  and  subtracted  in   the  same 
manner  as  compound  numbers. 

EXAMPLES    FOR   PRACTICE. 

1.  Add  12  ft.  7'  8",  15  ft.  3'  5",  17  ft.  9'  7". 

Am.  45  ft.  8'  8". 

2.  Add  136  ft.  11'  6"  8'",  145  ft.  10'  8"  5'",  160  ft,  9'  5"  5'", 

Ans.  443  ft.  V  8"  6'". 

3.  From  36  ft.  7'  11"  take  12  ft.  9'  5".      Ans.  23  ft.  10'  6". 

4.  A  certain  room  required  300  sq.  yd.  2  sq.  ft.  5'  of  plastering. 
The  walls  required  50  sq.  yd.  1  sq.  ft,  7'  4",  62  sq.  yd.  5'  3",  48 
sq.  yd.  2  sq.  ft.,  and  42  sq.  yd.  2  sq.  ft.  3'  4",  respectively.     Re- 
quired the  area  of  the  ceiling.      Ans.  97  sq.  yd.  5  sq.  ft.  1'  V. 


228  DUODECIMALS. 

MULTIPLICATION. 

38O.    In  the  multiplication  of  duodecimals,  the  product  of  two 
dimensions    is    area,    and    the   product   of  three    dimensions   is 
solidity  (£82,  286). 
We  observe  that 

V   X  1ft.  =  ^  of  1ft.  =1'. 

ix/  x  i  ft.  =  TiT  of  i  a    =  i". 

V    X  V      =  A  *  TL  of  1  ft.    =  I". 
\"  x  V     =  Th  X  3\  of  1  ft.  ==  1'".     Hence, 
The  product  of  any  two  orders  is  of  the  order  denoted  by  the 
sum  of  their  indices. 

39O.    1.  Multiply  9  ft.  8'  by  4  ft.  7'. 

OPERATION.  ANALYSIS.    Beginning  at  the  right, 

9  ft      g/  8'  X   7'  =  56"  =  ¥  8"  ;   writing 

4  ft      7'  the  S//  one  place  to  the  right,  we  re- 

—  —  —  -  —  —  —  serve  the  4X  to  be  added  to  the  next 

3°  «•      «,  product.     Then,  9  ft.  X  7'  +  4^  = 

_  I  --  67X  =  5  ft.  7/,  which  we  write  in  the 


44  ft.     3'     8",  Ans.          places  of  feet  and  primes.    Next  mul- 
tiplying by  4  ft.,  we  have  8X  X  4  ft. 

=  32/  =  2  ft.  8';  writing  the  8X  in  the  place  of  primes,  we  reserve  the 
2  ft.  to  be  added  to  the  next  product.  Then,  9  ft.  X  4  ft.  -f-  2  ft.  = 
38  ft.,  which  we  write  in  the  place  of  feet.  Adding  the  partial  pro- 
ducts, we  have  44  ft.  3X  8X/  for  the  product  required.  Hence  the 

RULE.     I.    Write  the  several  terms  of  the  multiplier  under  the 
corresponding  terms  of  the  multiplicand. 

II.  Multiply  each  term  of  the  multiplicand  by  each  term  of  the 
multiplier,  beginning  with  the  lowest  term  in  each,  and  call  the  pro- 
duct of  any  two  orders,  the  order  denoted  by  the  sum  of  their  in- 
dices, carrying  1  for  every  12. 

III.  Add  the  partial  products  ;  their  sum  will  be  the  required 
answer. 

EXAMPLES   FOR   PRACTICE. 

1.  How  many  square  feet  in  a  floor  16  ft.  8'  wide,  and  18  ft.  5f 
long? 

2.  How  much  wood  in  a  pile  4  ft.  wide,  3  ft.  8'  high,  and  23  ft 
TMong? 


MULTIPLICATION.  229 

3.  If  a  floor  be  79  ft.  8'  by  38  ft.  11',  how  many  square  yards 
does  it  contain  ?  Ans.  344  yd.  4  ft.  4'  4". 

4.  If  a  block  of  marble  be  7  ft.  6'  long,  3  ft.  3'  wide,  and  1  ft. 
10'  thick,  what  are  the  solid  contents?  Ans.  44  ft.  8'  3". 

5.  How  many  solid  feet  in  7  sticks  of  timber,  each  56  ft.  long, 
11  inches  wide,  and  10  inches  thick?  Ans.  299  ft.  5'  4". 

6.  How  many  feet  of  boards  will  it  require  to  inclose  a  building 
60  ft.  6'  long,  40  ft.  3'  wide,  22  ft.  high,  and  each  side  of  the 
roof  24  ft.  2',  allowing  523  ft.  3'  for  the  gables,  and  making  no 
deduction  for  doors  and  windows  ?  Ans.  7880  ft.  5; 

CONTRACTED   METHOD. 

391.  The  method  of  contracting  the  multiplication  of  deci- 
mals may  be  applied  to  duodecimals,  the  only  modification  being 
in  carrying  according  to  the  duodecimal,  instead  of  the  decimal, 
scale. 

1.  Multiply  7  ft.  3'  5"  8'"  by  2  ft,  4'  7"  9"',  rejecting  all  de- 
nominations below  seconds  in  the  product. 

OPERATION.  ANALYSIS.     We  write  2  ft.,  the 

7  ft.  3'  5"  8'"  units  of  the  multiplier,  under  the 

9'"  7"     4'  2  ft  lowest  order  to  be  reserved  in  the 

14  ft    6'  11"  product,  and  the  other  terms  at  the 

2ft.  5'     2"  left,  with  their  order  reversed.  Then 

4'     3"  it  is  obvious  that  the  product  of 

5"  each  term  by  the  one   above  it  is 

;jj~ffc J     9"-£-   ^nSf      seconds.     Hence  we  multiply  each 

term  of  the  multiplier  into  the  terms 

above  and  to  the  left  of  it  in  the  multiplicand,  carrying  from  tht, 
rejected  terms,  thus ;  in  multiplying  by  2  ft.,  we  have  8/x/  X  2  ft.  = 
16//x  =  1"  4//x,  which  being  nearer  1"  than  2X/,  gives  1"  to  be  car- 
ried to  the  first  contracted  product.  In  multiplying  by  4X,  we  have 
5//  x  4/  =  20'"  =  1"  8//x,  which  being  nearer  2/x  than  l/x,  gives 
2"  to  be  carried  to  the  second  contracted  product,  and  so  on. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  7  ft.  3'  4"  5'"  by  5  ft  8;  6",  extending  the  pro- 
duct only  to  primes.  Ans.  41  ft.  7'rfc 


230  DUODECIMALS. 

2.  How  many  yards  of  carpeting  will  cover  a  floor"  36  ft.  9'  4" 
long,  and  26  ft.  6'  9"  wide  ? 

3.  How  many  cu.  ft.  in  a  block  of  marble  measuring  6  ft.  2'  V 
in  length,  3  ft.  3'  4"  wide,  and  2  ft.  8'  6"  thick  ? 

4.  Find  the  product  of  7  ft.  6'  8",  3  ft.  2'  11",  and  3  ft.  8'  4", 
correct  to  within  1'.  Ans.  90  ft.  6f=t. 

DIVISION. 
392.   1.  Divide  41  ft.  8'  7"  6'"  by  7  ft.  5'. 

OPERATION.  ANALYSIS.       Divid- 

7  ft.  5')41  ft.  8'  1"  6"'(5  ft.  7'  6"          ing  the  units  of  the 

37  ft.  V  dividend  by  the  units 

4  ft   7'     7"  °^  ^e  divisor,  we  ob- 

4  ft.  3'  11"  tain  5  ft.  for  the  first 

^     7^7  „,„  term  of  the  quotient, 

3'     8"  6'"  anc*  ^  ^"  ^'  ^or  a  re~ 

mainder.        Bringing 

down  the  next  term  of  the  dividend,  we  have  4  ft.  7/  7//  for  a  new 
dividend.  Reducing  the  first  two  terms  to  primes,  we  have  55X  7/x>, 
whence  by  trial  division  we  obtain  7/  for  the  second  term  of  the  quo- 
tient, and  3X  8/x  for  a  remainder.  Completing  the  division  in  like 
manner,  we  have  5  ft.  7/  6/x  for  the  entire  quotient  Hence  the  fol- 
lowing 

RULE.     I.    Write  the  divisor  on  the  left  hand  of  the  dividend, 
as  in  simple  numbers. 

II.  find  the  first  term  of  the  quotient  either  by  dividing  the 
first  term  of  the  dividend  by  the  first  term  of  the  divisor,  or  by 
dividing  the  first  two  terms  of  the  dividend,  by  the  first  two  terms 
of  the  divisor ;  multiply  the  divisor  by  this  term  of  the  quotient, 
subtract  the  product  from  the  corresponding  terms  of  the  dividend, 
and  to  the  remainder  bring  down  another  term  of  the  dividend. 

III.  Proceed  in  like  manner  till  there  is  no  remainder,  or  till  a 
quotient  has  been  obtained  sufficiently  exact. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  287  ft.  7'  by  17  ft.  Ans.  16  ft.  11'. 

2.  Divide  29  ft.  5'  4"  by  6  ft.  8'.  Ans.  4  ft.  5'. 


DIVISION.  231 

3.  A  floor  whose  length  is  48  ft.  6'  has  an  area  of  1176  ft.  1' 
6" ;  what  is  its  width  ?  Ans.  24  ft.  3'. 

4.  From  a  cellar  38  ft.  10'  long  and  9  ft.  4'  deep,  were  exca- 
vated 275  cu.  yd.  5  cu.  ft.   V  4"  of  earth;  how  wide  was  the 
cellar  ?  Ans.  20  ft.  6'. 

CONTRACTED    METHOD. 

SOS.  Division  of  Duodecimals  may  be  abbreviated  after  the 
manner  of  contracted  division  of  decimals. 

1.  Divide  35  ft.  11'  11"  by  4  ft.  3'  7"  3'",  and  find  a  quotient 
correct  to  seconds. 

OPERATION. 

4  ft.  3'  7"  3'"  )  35  ft.  11'  11"  (  8  ft.  4'  5" 
34  ft.    4'  10" 


1  ft. 
1ft. 

r  i" 

5'     2" 

V  11" 

V    9" 

2",  rem. 

ANALYSIS.  Having  obtained  by  trial,  8  ft.  for  the  first  term  of  the 
quotient,  we  multiply  three  terms  of  the  divisor,  4  ft.  3'  7//,  carrying 
from  the  rejected  term,  t>'"  X  8  =  24//x  =  2/x,  making  34  ft.  4'  10XX, 
which  subtracted  from  the  dividend  leaves  1  ft.  7'  lxx  for  a  new  divi- 
dend. In  the  next  division,  we  reject  2  terms  from  the  right  of  the 
divisor,  and  at  the  last  division,  3  terms,  and  obtain  for  the  required 
quotient,  8  ft.  4X  5XX. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  7  ft.  7'  3"  by  2  ft.  10'  7",  extending  the  quotient  to 
seconds.  Ans.  2  ft.  7'  8"db. 

2.  Separate  64  ft.  9'  8"  into  three  factors,  the  first  and  second 
of  which  shall  be  7  ft.  2'  4"  and  4  ft.  7'  9"  8'"  respectively,  and 
obtain  the  third  factor  correct  to  within.  1  second. 

Ans.  1  ft.  11'  3"db. 

3.  What  is  the  width  of  a  room  whose  area  is  36  ft.  4'  8"  and 
whose  length  7  ft.  2'  11"  ? 


232  SHORT  METHODS 

< 

SHORT   METHODS. 

30  £.  Under  the  heads  of  Contractions  in  Multiplication  and 
Contractions  in  Division,  are  presented  only  such  short  methods 
as  are  of  the  most  extensive  application.  The  short  methods 
which  follow,  although  limited  in  their  application,  are  of  much 
value  in  computations. 

FOR   SUBTRACTION. 

£195.  When  the  minuend  consists  of  one  or  more 
digits  of  any  order  higher  than  the  highest  order  in  the 
subtrahend. 

The  difference  between  any  number  and  a  unit  of  the  next 
higher  order  is  called  an  Arithmetical  Complement.  Thus,  4  is  the 
arithmetical  complement  of  6,  31  of  69,  2792  of  7208,  etc. 

I.  Subtract  29876  from  400000. 

OPERATION.  ANALYSIS.     To  subtract  29876  from  400000  is  the 

400000          same  as  to  subtract  a  number  one  less  than  29876,  or 
29876          29875,  from  399999  (Ax.  2).     We  therefore  diminish 
3701^4          *^e  ^  °^  *ke  m^nuenc^  by"  1,  and  then  take  each  figure 
of  the  subtrahend  from  9,  except  the  last  or  right- 
hand  digit,  which  we  subtract  from  10.     Hence  the 

RULE.  I.  Subtract  1  from  the  significant  part  of  the  minuend 
and  write  the  remainder,  if  any-,  as  a  part  of  the  result. 

II.  Proceeding  Jo  the  right,  subtract  each  figure  in   the  subtra- 
hend from  9,  except  the  last   significant  figure,  which    subtract 
from  10. 

EXAMPLES    FOR   PRACTICE. 

1.  Subtract  756  from  1000.  Ans.  244. 

2.  Subtract  8576  from  4000000.  Ans.  3991424. 

3.  Subtract  .5768  from  10. 

4.  Subtract  13057  from  1700000. 

5.  Subtract  90.59876  from  64000. 

6.  Subtract  599948  from  1000000. 

7.  What  is  the  arithmetical  complement  of  271  ?     Of  18365  ? 
Of  3401250? 


FOR  MULTIPLICATION.  233 

FOR   MULTIPLICATION. 
CASE  I. 

396.  When  the  multiplier  is  9,  99,  or  any  number 
of  9's.  * 

Annexing  1  cipher  to  a  number  multiplies  it  by  10,  two  ciphers  by 
100,  three  ciphers  by  1000,  etc.  Since  9  is  10  —  1,  any  number  may 
be  multiplied  by  9  by  annexing  1  cipher  to  it  and  subtracting  the 
number  from  the  result.  For  similar  reasons,  100  times  a  number  —=•• 
1  time  tne  number  =  99  times  the  number,  etc.  Hence, 

RULE.  Annex  to  the  multiplicand  as  many  ciphers  as  the  multi* 
plier  contains  9's,  and  subtract  the  multiplicand  from  the  result. 

EXAMPLES    FOB   PRACTICE. 

1.  Multiply  784  by  99.  Ans.  77616. 

2.  Multiply  5873  by  .999. 

3.  Multiply  4783  by  99999.  Am.  478295217. 

4.  Multiply  75  by  999.999. 

CASE   II. 

397.  When  the  multiplier  is  a  number  a  few  units 
less  than  the  next  higher  unit. 

Were  we  required  to  multiply  by  97,  which  is  100  —  3,  we  could 
evidently  annex  2  ciphers  to  the  multiplicand,  and  subtract  3  times 
the  multiplicand  from  the  result.  Were  our  multiplier  991,  which  is 
1000  —  9,  we  could  subtract  9  times  the  multiplicand  from  1000  times 
the  multiplicand.  Hence, 

RULE.  I.  Multiply  by  the  next  higher  unit  by  annexing 
ciphers. 

II.  From  this  result  subtract  as  many  times  the  multiplicand 
as  there  are  units  in  the  difference  between  the  multiplier  and  th* 
next  higher  unit. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  786  by  98.  Ans.  77028. 

2.  Multiply  4327  by  96.  Ans.  415392. 

3.  Multiply  7328  by  997. 


234  SHORT  METHODS 

4.  Multiply  7873.586  by  9.95.  Ans.  78342.18070. 

5.  Multiply  43789  by  9994. 

6.  Multiply  7077364  by  .999993. 

CASE   ITI. 

398.  When  the  left  hand  figure  of  the  multiplier  is 
the  unit,  1,  the  right  hand  figure  is  any  digit  whatever, 
and  the  intervening  figures,  if  any,  are  ciphers. 

I.  Multiply  3684  by  17. 

OPERATION.  ANALYSIS.     If  we  multiply  by  the  usual 

3684  X   17  method,  we  obtain,  separately,  7  times  and 

g9g9g  10  times  the  multiplicand,  and  add  them. 

We  may  therefore  multiply  by  the  7  units, 

and  to  the  product  add  the  multiplicand  regarded  as  tens,  thus :  7  times 
4  is  28,  and  we  write  the  8  as  the  unit  figure  of  the  product.  Then, 
7  times  8  is  56,  and  the  2  reserved  being  added  is  58,  and  the  4  in 
the  multiplicand,  added,  is  62,  and  we  write  2  in  the  product.  Next, 
7  times  6,  plus  the  6  reserved,  plus  the  8  in  the  multiplicand,  is  56, 
and  we  write  6  in  the  product.  Next,  7  times  3,  plus  the  5  reserved, 
plus  the  36  in  the  multiplicand,  is  62,  which  we  write  in  the  product, 
and  the  work  is  done. 

Had  the  multiplier  been  107,  we  should  have  multiplied  two  figures 
of  the  multiplicand  by  7,  before  we  commenced  adding  the  digits  of 
the  multiplicand  to  the  partial  products ;  3  figures  had  the  multiplier 
been  1007,  etc".  Hence  the 

RULE.  I.  Write  the  multiplier  at  the  right  of  the  multiplicand, 
with  the  sign  of  multiplication  between  them. 

II.  Multiply  the  multiplicand  by  the  unit  Jiyure  of  the  multi- 
plier, and  to  the  product  add  the  multiplicand,  regarding  its  local 
value  as  a  product  l>y  the  left  hand  figure  of  the  multiplier. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  567  by  13.  Ans.  7371. 

2.  Multiply  439603  by  10.5.  Ans.  4615831.5. 

3.  Multiply  7859  by  107. 

4.  Multiply  18075  by  1008.  Ans.  18219600. 

5.  Multiply  3907  by  10.002. 


FOR  MULTIPLICATION.  235 

CASE  IV. 

399.  When  the  left  hand  figure  of  the  multiplier  is 
any  digit,  the  right  hand  figure  is  the  unit,  1,  and  the 
intermediate  figures,  if  any,  are  ciphers. 

I.  Multiply  834267  by  301. 

OPERATION.  ANALYSIS.     Regarding  the  multipli- 

834267  X  301          ca,nd  as  a  product  by  the  unit,  1,  of  thef 

7-  multiplier,  we   multiply  the   multipli-' 

14367  cand  by  3  hundreds,  and  add  the  digits 

of  the  multiplicand  to  the  several  products  as  we  proceed.  Since  the 
3  is  hundreds,  the  two  right  hand  figures  of  the  multiplicand  will 
be  the  two  right  hand  figures  of  the  product ;  and  the  product  of  3  X  7 
will  be  increased  by  2,  the  hundreds  of  the  multiplicand. 

Had  the  multiplier  been  31,  the  tens  of  the  multiplicand  would 
have  been  added  to  3  X  7  ;  had  the  multiplier  been  3001  the  thousands 
of  the  multiplicand  would  have  been  added  to  3  X  7 ;  and  so  on. 
Hence  the 

RULE.  I.  Write  the  multiplier  at  the  right  of  the  multiplicand, 
with  the  sign  of  multiplication  between  them. 

II.  Multiply  the  multiplicand  l>y  the  left  hand  figure  of  the  mul- 
tiplier, and  to  the  product  acid  the  multiplicand,  regarding  its  local 
value  as  a  product  by  the  unit  figure  of  the  multiplier. 

EXAMPLES   FOR   PEACTICE. 

1.  Multiply  56783  by  71. 

2.  Multiply  47. 89  by  60.1.  Ans.  2878.189. 

3.  Multiply  3724.5  by  .901 

4.  Multiply  103078  by  40001.  Ans.  4123223078. 

CASE    V. 

400.  When  the  digits  of  the  multiplier  are  all  the 
same  figure. 

1.  Multiply  81362  by  333. 

OPERATION.  ANALYSIS.     We  first  multiply  by  999,  by 

81362000          (396).    Then,  since  333  is  £  of  999,  we  take 

81362          i  of  the  product, 

o  -\  Qi9er)fQQ  Had  our  multiplier  been  444,  we  would 

have  taken  £  of  999  times  the  multiplicand. 
27093546          Had  it  been  66,  we  would  have  taken  £  =  f 
of  99  times  the  multiplicand,  etc.     Hence 


236  SHORT  METHODS 

RULE  I.  Multiply  by  as  many  9's  as  the  multiplier  contains 
digits,  by  (396). 

II.  Take  such  a  part  of  the  product  as  1  digit  of  the  multiplier 
is  part  of  9. 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  432711  by  222.  Ans.  96061842. 

2.  Multiply  578  by  1111. 

3.  Multiply  .6732  by  88.888.  Ans.  59.8394016. 

4.  Multiply  8675  by  77.7. 

5.  Multiply  44444  by  88888. 


CASE   VI. 

4O1.  To  square  a  number  consisting  of  only  two 
digits. 

I.  What  is  the  square  of  18  ? 
ANALYSIS.     According  to  (86),  we  have 

182  =  18  X  18 

Now  if  one  of  these  factors  be  diminished  by  2,  the  product  will  be 
less  than  the  square  of  18  by  2  times  the  other  factor,  (93, 1) ;  that  is, 

182=(16  X  18) +  (2  x  18). 

Next,  if  we  increase  the  other  factor,  18,  in  this  result,  by  2,  the 
whole  result  will  exceed  the  square  of  18,  by  2  times  the  other  factor, 
16,  (93,  III);  that  is, 

182  =  (16  X  20)  +  (2  x  18)  —  (2  X  16). 
But  as  2  times  18  minus  2  times  16  is  equal  to  2  X  2,  or  22,  we 

have 

182  =  16  X  20  +  22.     Hence  the 

RULE.  I.  Take  two  numbers,  one  of  which  is  as  many  units 
less  than  the  number  to  be  squared  as  the  other  is  units  greater,  and 
one  of  the  numbers  taken  an  exact  number  of  tens. 

II.  Multiply  these  two  numbers  together,  and  to  the  product  add 
the  square  of  the  difference  between  the  given  and  one  of  the  as- 
sumed numbers. 

NOTE.  —  A  little  practice  will  enable  the  pupil  to  readily  square  any  number 
less  than  100  mentally  by  this  rule. 


.    FOR  MULTIPLICATION.  237 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  square  of  27  ?  Ans.  729. 

2.  What  is  the  square  of  49  ?  Ans.  2401. 

3.  Square  28,  26,  39,  38,  37,  36,  and  35. 

4.  Square  77,  88,  8.6,  99,  98,  69,  68,  6.7,  and  62. 

CASE    VII. 

402.  When  the  multiplier  is  an  aliquot  part  of  some 
higher  unit. 

An  Aliquot  or  Even  Part  of  a  number  is  such  a  part  as  will 
exactly  divide  that  number.  Thus,  5,  8£,  and  12£  are  aliquot 
parts  of  25  and  of  100,  etc. 

NOTE. — An  aliquot  port  may  be  either  a  whole  or  a  mixed  number,  while  n 
component  factor  must  be  a  whole  number. 

403.  The  aliquot  parts  of  10  are  5,  3|,  2-J,  2,  If,  If,  1-j,  1£. 
The  aliquot  parts  of  100,  1000,  or  of  any  other  number,  may 

be  found  by  dividing  the  number  by  2,   3,    4,    etc.,    until  it  has 
been  divided  by  all  the  integral  numbers  between  1  and  itself. 

I.  Multiply  78  by  3 i,  and  by  25  separately. 

OPERATION.  ANALYSIS.     Since  3£  is  £  of  10, 

3  )  780  4  )  7800          the  next  higher  unit,  we  multiply 

~260    an<*          1950          ^8  bv  10  and  take  ^  of  the  Pr°duct. 
Again,  since  25  is  \  of  100,  we 
multiply  78  by  100  and  take  \  of  the  product.     Hence  the 

E-ULE.  I.  Multiply  the  given  multiplicand  ~by  the  unit  next 
higher  than  the  multiplier,  by  annexing  ciphers. 

II.  Take  such  a  part  of  this  product  as  the  given  multiplier  is 
part  of  the  next  higher  unit. 

EXAMPLES    FOR   PRACTICE. 

1.  Multiply  437  by  25.  Ans.  10925. 

2.  Multiply  6872  by  2£.  Ans.  17180. 

3.  Multiply  5734154  by  333f  Ans.  1911384666f . 

4.  Multiply  758642  by  12J. 

5.  Multiply  78563  by  125.  Ans.  9820375. 

6.  Multiply  57687  by  142f 


238  SHORT  METHODS 

CASE   VIII. 

404.  When  the  right  hand  figure  or  figures  of  the 
multiplier  are  aliquot  parts  of  10,  100,  1000,  etc. 

I.  Multiply  2183  by  1233£. 

OPERATION. 

218300 

12*  ANALYSIS.     1233£  =  12$  X  100.    We  there- 

797661  f°re  multiP1y  ky  100'  and  by  12^'  in  continued 

^6196  multiplication.     Hence  the 

"26923661 

RULE.  I.  Reject  from  the  right  hand  of  the  multiplier  such 
figure  or  figures  as  are  an  aliquot  part  of  some  higher  unit,  and 
to  the  remaining  figures  of  the  multiplier  annex  a  fraction  ivhich 
expresses  the  aliquot  part  thus  rejected,  for  a  reserved  multiplier. 

II.  Annex  to  the  multiplicand  as  many  ciphers  as  ore  equal  to 
the  number  of  figures  rejected  from  the  right  hand  of  the  multi- 
plier, and  multiply  the  result  by  the  reserved  multiplier. 

EXAMPLES   FOR   PRACTICE. 

1.  Multiply  43789  by  825.  Ans.  36125925. 

2.  Multiply  58730  by  7125. 

3.  Multiply  7854  by  34.2*.  Ans.  268999.5. 

4.  Multiply  30724  by  73333$. 

5.  Multiply  47836  by  712*.  Ans.  34083150. 

6.  Multiply  53727  by  2416f  . 

CASE   IX. 

405.  To  find  the  cost  of  a  quantity  when  the  price 
is  an  aliquot  part  of  a  dollar. 

1.  What  cost  a  case  of  muslins  containing  1627  yds.,  at  $.12* 
per  yard  ? 

OPERATION.  ANALYSIS.    At  $1  per  yard  the  case  would 

8  )  $1627  cost  as  many  dollars  as  it  contained  yards  ; 


$203  37*          and  at    '12   =      per  yard'  Jt  would  cost  i 
as  many  dollars  as  it  contained  yards.     We 

therefore  regard  the  yards  as  dollars,  which  we  divide  by  8.     Hence, 


.  FOR  MULTIPLICATION.  239 

RULE.  Take  such  a  part  of  the  given  quantity  as  the  price  is 
part  of  one  dollar. 

NOTK.  —  Since  the  shilling  in  most  of  the  different  currencies  is  some  aliquot 
part  of  the  dollar,  this  rule  is  of  much  practical  use  in  making  out  bills  and 
accounts  where  the  prices  of  the  items  are  given  in  State  Currency,  and  the 
amounts  are  required  in  United  States  Money. 

EXAMPLES   FOR   PRACTICE. 

]  .  What  cost  568  pounds  of  butter  at  25  cents  a  pound  ? 

Ans.  SI  42. 

2.  A  merchant  sold  51  yards  of  prints  at  16f  cents  per  yard,  8 
pieces  of  sheeting,  each  piece  containing  33  yards,  at  6i  cents  per 
yard,  and  received  in  payment  18  bushels  of  oats  at  33£  cents  per 
bushel,  and  the  balance  in  money  ;  how  much  money  did  he  re- 
ceive ?  Ans.  $19.    » 

3.  Required  the  cost  of  28   dozen  candles,  at  1  shilling  pei 
dozen,  New  York  currency.  Ans.  $3.50. 

4.  What  cost  576  Ibs.  of  beef  at  lOd.  per  pound,  Pennsylvania 
currency?  Ans.  $64. 

5.  If  a  grocer  in  New  York  gain  $7.875  on  a  hogshead  of  mo- 
lasses containing  63  gallons,  how  much  will  he  gain  on  576  gallons 
at  the  same  rate  ?  Ans.  $72. 

CASE  x. 

4O6.  To  find  the  cost  of  a  quantity,  when  the  quan- 
tity is  a  compound  number,  some  part  or  all  of  which  is 
an  aliquot  part  of  the  unit  of  price. 

1.  What  cost  5  bu.  3  pk.  4  qt.  of  cloverseed,  at  $3.50  per  bu.  ? 

OPERATION.  ANALYSIS.     Multiply- 

2,,   4,,   8  )  $3.50  price.  ing  the  price  'by  5,  we 

5  have  the  cost  of  5   bu. 

$17.50    cost  of  5  bu.  Dividing  the  price  by  2, 

1.75       "     "  2  pk  we  ^ave  t*10  cost  of  J  bu. 

.875     "     "  1    "  =2   pk.     Dividing   the 

.4375  "     "  4  qt.  price  by  4,  or  the  cost  of 


$205625,  Ans.  2  Pk'  ky  2'  7e  l£v?  *he 

cost  of  1   pk.     Dividing 

the  price  by  8,  or  the  cost  of  2  pk.  by  4,  or  the  cost  of  1  pk.  by  2,  we 


240  SHORT  METHODS 

have  the  cost  of  £  pk.  =  4  qt.  And  the  sum  of  these,-  several  values 
is  the  entire  cost  required. 

2.  At  £6  7s.  5£d.  Sterling  per  hhd.,  how  much  will  4  hhd.  9 
gal.  3  qt.  of  West  India  Molasses  cost  ? 

OPERATION.  ANALYSIS.     Mul- 

7)  £6     7s.       5£d.  price.  tiplying    the    price 

4  by  4,  we  have  the 

"  25     9  «    10  «         cost  of  4  hhd.      cost  of  4  hhd-    Di- 

12)  "          18  "      2   "  2  qr.  "    "  9  gal.         viding  the  price  by 

I  «      Q  tt      6  (t  (t   a  3  qk          7,  We  have  the  cost 

«  26     9  «      6  "  2|, "  4ws.  of  4  hhd-  =  9  Sal' 

Dividing  the  cost  of 

9  gal.  by  12,  we  have  the  cost  of  -?•$  of  36  qt.  =  3  qt.     And  the  sum 
of  these  several  results  is  the  entire  cost  required. 
From  these  illustrations  we  deduce  the  following 

RULE.  I.  Multiply  the  price  Ly  the  number  of  units  of  the  de- 
nomination corresponding  to  the  price. 

II.  For  the  lower  denominations,  take  aliquot  parts  of  the  price  ; 
the  sum  of  the  several  results  will  be  the  entire  cost. 

NOTE. — This  method  is  applicable  in  certain  cases  of  multiplication,  where  one 
compound  number  is  taken  as  many  times  as  there  are  units  and  parts  of  a 
unit  of  a  certain  kind,  in  another  compound  number.  This  will  be  seen  in  the 
first  example  below. 

EXAMPLES    FOR   PRACTICE. 

1.  A  chemist  filtered  18  gal.  3  qt.  1  pt.  of  rain-water  in  1  day; 
at  the  same  rate  how  much  could  he  filter  in  4  da.  6  h.  30  min.  ? 
OPERATION.  ANALYSIS.     Multi- 

18  gal.  3  qt.  1  pt.  in  1  da.  Plv^g    the    quantity 

4  filtered  in  1  day  by  4, 

75    «     o  «    0  «  «   4    «  WG  haVG  the  <luantity 

4    «     2«     1«3     ri    «    6h  filtered  in  4  day,  Di- 

1M     1«     TV<'«30mm.  viding  the  quantity  fil- 

— • 12  tered  m  1  day  by  4, 

80    «      2   «     0  «  3/3  «     Ans.  we  have  the  quantity 

filtered  in  \  da.  =  6 

h.  Dividing  the  quantity  filtered  in  6  hours  by  12,  we  have  the  quan- 
tity filtered  in  £  h.  =  30  min.  And  the  sum  of  these  several  results 
is  the  entire  result  required. 


FOR  DIVISION.  241 

2.  What  will  be  the  cost  of  3  Ib.  10  oz.  8  pwt.  5£  gr.  of  gold 
at  $15.46  per  oz.  ?  Ans.  $717.52. 

8.  A  man  bought  5  cwt.  90  Ib.  of  hay  at  $.56  per  cwt. ;  what 
was  the  cost?  Ans.  $3.304. 

4.  AY  hat  must  be  given  for  3  bu.  1  pk.  3  qt.  of  cloverseed,  at 
$4.48  per  bushel?  Ans.  $14.98. 

5.  A  gallon  of  distilled   water  weighs  8  Ib.  5.42  oz. ;  required 
the  weight  of  5  gal.  3  qt.  1  pt.  3  gi. 

Ans.   49  Ib.l2.3506j-  oz. 

6.  At  $17.50  an  acre,  what  will  3  A.  1  R.  35.4  P.  of  land  cost? 

7.  If  an  ounce  of  English  standard  gold  be  worth  £3  17s.  10|d., 
what  will  be  the  value  of  an  ingot  weighing  7  oz.  16  pwt.  18  gr.  ? 

Ans.  £30  10s.  4.14375d, 

8.  If  a  comet  move  through  an  arc  of  4°  36'  40"  in  1  day,  how 
far  will  it  move  in  5  da.  15  h.  32  min.  55  sec.  ? 

9.  What  will  be  the  cost  of  7  gal.  1  qt.  1  pt.  3  gi.  of  burning 
fluid,  at  4s.  8d.  per  gallon,  N.  Y.  currency?         Ans.  $4.35-f . 

10.  What  must  be  paid  for  12  J  days'  labor,  at  5s.  3d.  per  day, 
New  England  currency  ? 

FOR  DIVISION. 
CASE   I. 

407.  When  the  divisor  is  an  aliquot  part  of  some 
higher  unit. 

1.  Divide  260  by  3J,  and  1950  by  25. 

OPERATION.  ANALYSIS.     Since  3£  is  £  of  10,  the  next 

26 1 0          19 1 50          higher  unit,  we  divide  2GO  by  10  ;  and  hav- 

3     and    4  ing  used  3  times  our  true  divisor,  we  obtain 

i^ ^  only  £  of  our  true  quotient.      Multiplying 

the  result,  26,  by  3,  we  have  78,  the  true 

quotient.  Again,  since  25  is  \  of  100,  the  next  higher  unit,  we  divide 
1950  by  100 ;  and  having  used  4  times  our  true  divisor,  the  result, 
19.5,  is  only  \  of  our  true  quotient.  Multiplying  19.5  by  4,  we  have 
78,  the  true  quotient.  Hence  the 

RULE.  I.  Divide  the  given  dividend  by  a  unit  of  the  order 
next  higher  than  the  divisor,  l>y  cutting  off  Iqures  from  the  right. 

21  Q 


242  SHORT  METHODS. 

II.    Take  as  many  times  this  quotient  as  the  divisor  is  contained 
times  in  the  next  hiyher  unit. 

EXAMPLES    FOB   PRACTICE. 

1.  Divide  63475  by  25.  Ans.  2539. 

2.  Divide  7856  by  1.25.  Ans.  6284.S. 

3.  Divide  516  by  33.3£. 

4.  Divide  16.7324  by  12i. 

5.  Divide  1748  by  .14f.  Ans.  12236 

6.  Divide  576.34  by  1.6§. 

CASE   II. 

4O8.  When  the  right  hand  figure  or  figures  of  the 
divisor  are  an  aliquot  part  of  10,  100,  1000,  etc. 
1.  Divide  26923661  by  1233*. 


OPERATION. 

$  ot  100,  we  multiply  both 
1233*  )  26923661  dividend  and  divisP0/by  3> 

(117,  HI),  and  we  obtain  a 

37|00  )  80771 10'0  (  2183,  Ans.          divisor  the  component  fac- 
67  tors  of  which  are  100  and  37. 

307  We   then  divide   after  the 

manner  of  contracted  divi- 
sion, (112). 
2.  Divide  601387  by  1875. 

OPERATION.  ANALYSIS.       Multiplying    both 

1875)    601387  dividend  and  divisor  by  4,  we  ob- 

4  4  tain  a  new  divisor,  7500,  having  2 

7500  'i  2405548  ciphers  on  the  right  of  it.     Multi- 

plying again  by  4,  we  obtain  a  new 
divisor,  30000,  having  4  ciphers  on 

3|0000)962|2192  the  right.     Then  dividing  the  new 

320i|f  I,  Ans.        dividend  by  the  new  divisor,  we  ob^ 

tain  320  for  a  quotient,  and  22192 

for  a  remainder.  As  this  remainder  is  a  part  of  the  new  dividend, 
it  must  be  4  X  4  =  16  times  the  true  remainder;  we  therefore  divide 
it  by  16,  and  write  the  result  over  the  given  divisor,  1875,  and  anneJ 
the  fraction  thus  formed  to  the  integers  of  the  quotient. 


RATIO.  243 

From  these  illustrations  we  derive  the  following 

RULE.      I.   Multiply  both  dividend  and  divisor  by  a  number  or 

numbers  that  will  produce  for  a  new  divisor  a  number  ending  in  a 

cipher  or  ciphers. 

II.   Divide  the  new  dividend  by  the  new  divisor. 

NOTE. — If  the  divisor  be  a  whole  number,  or  a  finite  decimal,  the  multiplier 
will  be  2,  4,  5,  or  8,  or  some  multiple  of  one  of  these  numbers. 

EXAMPLES    FOR   PRACTICE. 

1.  Divide  64375  by  2575. 

2.  Divide  76394  by  3625.  Ant.  21,2ff%- 
8.  Divide  7325  by  433i 

4.  Divide  5736  by  431.25.  Ant.  13^f. 

5.  Divide  42.75  by  566|. 

6.  Divide  24409375  by  .21875. 

7.  Divide  785  by  3.14f.  Ans.  249^. 


RATIO. 

Ratio  is  the  relation  of  two  like  numbers  with  respect 
to  comparative  value. 

NOTE.  —  There  are  two  methods  of  comparing  numbers  with  respect  to  value; 
1st,  by  subtracting  one  from  the  other;  2d,  by  dividing  one  by  the  other.  The 
relation  expressed  by  the  difference  is  sometimes  called  Arithmetical  Ratio,  and 
the  relation  expressed  by  the  quotient,  Geometrical  Itatio. 

410.  When  one  number  is  compared  with  another,  as  4  with 
12,  by  means  of  division,  thus,  12-^4  =  3,  the  quotient,  3,  shows 
the  relative  value  of  the  dividend  when  the  divisor  is  considered 
as  a  unit  or  standard.     The  ratio  in  this  case  shows  that  12  is  3 
times  4;  that  is,  if  4  be  regarded  as  a  unit,  12  will  be  3  units,  or 
the  relation  of  4  to  12  is  that  of  1  to  3. 

411.  Ratio  is  indicated  in  two  ways : 

1st.  By  placing  two  points  between  the  two  numbers  compared, 
writing  the  divisor  before  and  the  dividend  after  the  points,. 
Thus,  the  ratio  of  8  to  24  is  written  8  :  24;  the  ratio  of  7  to  5  is 
written  7  :  5. 


244  RATIO. 

2d.  In  the  form  of  a  fraction.  Thus,  the  ratio  of  8  to  24  is 
written  2g4  ;  the  ratio  of  7  to  5  is  f. 

4LI£$.    The  Terms  of  a  ratio  are  the  two  numbers  compared. 

The  Antecedent  is  the  first  term;  and 

The  Consequent  is  the  second  term. 

The  two  terms  of  a  ratio  taken  together  arc  called  a  couplet. 

418.    A  Simple  Ratio  consists  of  a  single  couplet;  as  5  :  15. 

414.  A  Compound  Ratio  is  the  product  of  two  or  more  sim- 
ple ratios.  Thus,  from  the  two  simple  ratios,  5  :  1G  and  8  :  2,  we 

5  :  16 
8  :    2 
may  form  the  compound  ratio  5^8  :  10x~2,  or  ^  X  §  =  ^'2  =  4. 

41*5.  The  Reciprocal  of  a  ratio  is  1  divided  by  the  ratio;  or, 
which  is  the  same  thing,  it  is  the  antecedent  divided  by  the  con- 
sequent. Thus,  the  ratio  of  7  to  9  is  7  :  9  or  |,  and  its  reciprocal 


NOTK.  —  The  quotient  of  the  second  term  divided  by  the  fir?t  is  sometimes 
called  a  Direct  R«tu>,  and  the  quotient  of  the  first  term  divided  by  the  second, 
fin  In  Verne  or  Reciprocal  Ratio. 

4IO.  One  quantity  is  said  to  vary  directly  as  another,  when 
the  two  increase  or  decrease  together  in  the  same  ratio  ;  and  one 
quantity  is  said  to  vary  inversely  as  another,  when  one  increases 
in  the  same  ratio  as  the  other  decreases.  Thus  time  varies  directly 
as  wages  ;  that  is,  the  greater  the  time  the  greater  the  wages,  and 
the  less  the  time  the  less  the  wages.  Again,  velocity  varies  in- 
versely as  the  time,  the  distance  being  fixed;  that  is,  in  traveling 
a  given  distance,  the  greater  the  velocity  the  less  the  time,  and  the 
less  the  velocity  the  greater  the  time. 

41T.  Ratio  can  exist  only  between  like  numbers,  or  between 
two  quantities  of  the.  same  kind.  But  of  two  unlike  numbers  or 
quantities,  one  may  vary  either  directly  or  inversely  as  the  other. 
Thus,  cost  varies  directly  as  quantity,  in  the  purchase  of  goods: 
time  varies  inversely  as  velocity,  in  the  descent  of  falling  bodies. 
In  all  cases  of  this  kind,  the  quantities,  though  unlike  in  kind, 
have  a  mutual  dependence,  or  sustain  to  each  other  the  relation 
of  cause  and  effect. 


RATIO.  245 

418.  In  the  comparison  of  like  numbers  we  observe, 

I.  If  the  numbers  are   simple,  whether  abstract  or   concrete, 
their  ratio  may  be  found  directly  by  division. 

II.  If  the  numbers  are  compound,  they  must  first  be  reduced 
to  the  same  unit  or  denomination. 

III.  If  the  numbers  are  fractional,  and  have  a  common  de- 
nominator, the  fractions  will  be  to  each  other  as  their  numerators ; 
if  they  have  not  a  common  denominator,  their  ratio  may  be  found 
either  directly  by  division,  or  by  reducing  them  to  a  common 
denominator  and  comparing  their  numerators. 

419.  Since  the  antecedent  is  a  divisor  and  the  consequent  a 
dividend,  any  change  in  either  or  both  terms  will  be  governed  by 
the  general  principles  of  division,  (117)-     "We  have  only  to  sub- 
stitute  the   terms   antecedent,   consequent,  and  ratio,  for  divitor, 
dividend,  and  quotient,  and  these  principles  become 

GENERAL    PRINCIPLES    OF   RATIO. 

PRIN.  I.  Multiplying  the  consequent  multiplies  the  ratio ;  divi- 
ding the  consequent  divides  ihe  ratio. 

PRIN.  II.  Multiplying  the  antecedent  divides  the  ratio;  dividing 
the  antecedent  multiplies  the  ratio. 

PRIN.  III.  Multiplying  or  dividing  Loth  antecedent  and  cnnse-* 
quent  l>y  the  same  number  docs  not  alter  the  ratio. 

4SG.    These  three  principles  may  be  embraced  in  one 

GENERAL   LAW. 

A  change  in  the  consequent  by  multiplication  or  division  prodi*- 
ce*  a  LIKE  change  in  the  ratio  ;  but  a  change  in  the  antecedent 
pr'td  iii-ex  nn  OPPOSITE  change  in  the  ratio. 

431,  Since  the  ratio  of  two  numbers  is  equal  to  the  conse- 
quent divided  by  the  antecedent,  it  follows,  that 

I.  The  antecedent  is  equal  to  the  consequent  divided  by  the 
ratio;  and  that, 

II.  The  consequent  is  equal  to  the  antecedent  multiplied  by  the 
ratio. 

21* 


246  RATIO. 

EXAMPLES  FOR  PRACTICE. 

1.  What  part  of  28  is  7  ? 

2s  =  i  ;  or,  28  :  7  as  1  :  £ ;  that  is,  28  has  the  same  ratio  to  7  that 

1  has  to  \.  Ans.   £. 

2.  What  part  of  42  is  6? 

3.  What  is  the  ratio  of  120  to  80  ?  Am.  f . 

4.  What  is  the  ratio  of  8|  to  60  ?  Ans.  7. 

5.  What  is  the  ratio  of  T\  to  26  ? 

6.  What  is  the  ratio  of  7^  to  21?  Ans.  f$. 

7.  What  is  the  ratio  of  J  to  y7^  ?  Ans.  4i. 

8.  What  is  the  ratio  of  1  mi.  to  120  rd.?  Ans.  -|. 

9.  What  is  the  ratio  of  1  wk.  3  da.  12  h.  to  9  wk.?     Ans.  6. 

10.  What  is  the  ratio  of  10  A.  60  P.  to  6  A.  110  P.? 

11.  What  is  the  ratio  of  25  bu.  2  pk.  6  qt.  to  40  bu.  4.5  pk.  ? 

12.  What  is  the  ratio  of  18f  °  to  45'  30"  ? 

12£        ?-  Of  -3 

13.  What  part  of  -^-  is  ^-4-  ?  Ans.  Tf  5. 


14.  What  is  the  ratio  of  —    to  f  of  T%  of      ? 

s 

15.  Find  the  reciprocal  of  the  ratio  of  42  to  28.        Ans.  1^. 

16.  Find  the  reciprocal  of  the  ratio  of  3  qt.  to  43  gal. 

17.  If  the  antecedent  be  15  and  the  ratio  |,  what  is  the  conse- 
quent? Ans.  12. 

18.  If  the  consequent  be  3^  and  the  ratio  7,  what  is  the  ante- 
cedent? Ans.  1|. 

19.  If  the  antecedent  be  i  of  |  and  the  consequent  .75,  what 
is  the  ratio  ? 

20..  If  the  consequent  be  $6.12£  and  the  ratio  25,  what  is  the 
antecedent?  Ans.  $.245. 

21.  If  the  ratio  be  J  and  the  antecedent  |,  what  is  the  conse- 
quent ? 

22.  If  the  antecedent  be  13  A.  145  P.  and  the  ratio  |f,  v/liat 
is  the  consequent?  Ans.  6  A.  90  P. 


PROPERTIES  OF  PROPORTION.  247 


PROPORTION. 

422.  Proportion  is  an  equality  of  ratios.     Thus,  the  ratios 
5  :  10  and  6  :  12,  each  being  equal  to  2,  form  a  proportion. 

NOTE.  —  When  four  numbers  form  a  proportion,  they  are  said  to  be  propor- 
tional. 

423.  Proportion  is  indicated  in  three  ways : 

1st,  By  a  double  colon  placed  between  the  two  ratios;  thus, 
S  :  4  :  :  9  : 12  expresses  the  proportion  between  the  numbers  3,  4, 
9,  and  12,  arid  is  read,  3  is  to  4  as  9  is  to  12. 

2d.  By  the  sign  of  equality  placed  between  two  ratios;  thus, 
3  : 4  =  9  :  12  expresses  proportion,  and  may  be  read  as  above,  or, 
the  ratio  of  3  to  4  equals  the  ratio  of  9  to  12. 

3d.  By  employing  the  second  method  of  indicating  ratio;  thus, 
|  =  *-£  indicates  proportion,  and  may  be  read  as  either  of  the 
above  forms. 

424.  Since  each  ratio  consists  of  two  terms,  every  proportion 
must  consist  of  at  least  four  terms.     Of  these 

The  Extremes  are  the  first  and  fourth  terms;  and 

The  Means  are  the  second  and  third,  terms. 

423.  Three  numbers  are  proportional  when  the  first  is  to  the 
second  as  the  second  is  to  the  third.  Thus,  the  numbers  4,  6, 
and  9  are  proportional,  since  4:6=6:9,  the  ratio  of  each  couplet 
being  |,  or  1J, 

423.  When  three  numbers  are  proportional,  the  second  term 
is  called  the  Mean  Proportional  between  the  other  two. 

42*7.    If  we  have  any  proportion,  as 
3  :  15  ==  4  :  20, 

Then,  indicating  this  ratio  by  the  second  method,  we  have 

V  =  V- 

Reducing  these  fractions  to  a  common  denominator, 
15  X  4  =  20  X  3 
~!2~  12 

And  since  these  two  equal  fractions  have  the  same  denominator, 
the  numerator  of  the  first,  which  is  the  product  of  the  means,  must 
be  equal  to  the  numerator  of  the  second,  which  is  the  product  of  the 
extremes  ;  or,  15  X  4  =  20  X  3.  Hence, 


248  PROPORTION. 

I.  In  every  proportion  the  product  of  the  means  equals  the 
product  of  the  extremes. 

4.gain,  take  any  three  terms  in  proportion,  as 

4  :  6^=6  :  9 

Then,  since  the  product  of  the  means  equals  the  product  of  the  ex- 
tremes, 

62  =  4  x  9.     Hence, 

II.  The  square  of  a  mean  proportional  is  equal  to  the  product 
of  the  other  two  terms. 

4S8.  Since  in  every  proportion  the  product  of  the  means 
equals  the  product  of  the  extremes,  (4^57, 1),  it  follows  that,  any 
three  terms  of  a  proportion  being  given,  the  fourth  may  be  found 
by  the  following 

RULE.  I.  Divide  the  product  of  the  extremes  ~by  one  of  the 
means,  and  the  quotient  will  be  the  other  mean.  Or, 

II.  Divide  the  product  of  the  means  l>y  one  of  the  extremes}  and 
the  quotient  will  be  the  other  extreme. 

EXAMPLES    FOR   PRACTICE. 

The  required  term  in  an  operation  will  be  denoted  by  (?), 
which  may  be  read  "  how  many,"  or  "  how  much." 

Find  the  term  not  given  in  each  of  the  following  proportions : 

1.  4  :26  =  10  :  (?).  Ans.  65. 

2.  $8865  :  $720  =  (  ?  )  :  16  A.          Ans.  197  A. 

3.  4£  yd.  :(?)::  $9.75  :  $29.25.         Ans.  13£  yd. 

4.  (?)  :  21  A.  140  P.  :  :  $1260  :  $750. 

Ans.  36  A.  120  P. 

5.  7>50:18  =  (?):7TVoz. 

6.  7  oz  :(?)::  £30  :  £407  2s.  lOfd.        Ans.  7  Ib.  11  oz. 

7.  (?  )  :  .15  hhd.  :  :  $2.39  :  $.3585.  Ans.  1  hhd. 

8.  1  T.  7  cwt.  3  qr.  20  Ib.  :  13  T.  5  cwt.  2  qr.  =  $9.50  :  (  ?) 

9.  $175.35  :(?)  =  £:  f  Ans-  $601.20. 
10.  (?)  :  812£  =  240|  :  149TT72V          Ans-  $2°g- 
11-  f  yd.  :(?)::  $i  :  $59.0625.          Ans.  40^  yd. 


SIMPLE  PROPORTION. 


249 


CAUSE    AND    EFFECT. 

42Q.  Every  question  in  proportion  may  be  considered  as  a 
comparison  of  two  causes  and  two  effects.  Thus,  if  3  dollars  as 
o.  cause  will  buy  12  pounds  as  an  effect,  6  dollars  as  a  cause  will 
ouy  24  pounds  as  an  effect.  Or,  if  5  horses  as  a  cause  consume 
10  tons  as  an  effect,  15  horses  as  a  cause  will  consume  30  tons  as 
an  effect. 

Causes  and  effects  in  proportion  are  of  two  kinds  —  simple  and 
compound. 

430.  A  Simple  Cause  or  Effect  contains  but  one  element; 
as  price,  quantity,  cost,  time,  distance,  or  any  single  factor  used 
as  a  term  in  proportion. 

431.  A  Compound  Cause  or  Effect  is  the  product  of  two  or 
more  elements;  as  the  number  of  workmen  taken  in  connection 
with  the  time  employed,  length  taken  in  connection  with  breadth 
and   depth,   capital   considered   with   reference   to  the  time   cm- 
ployed,  etc. 

433.  Since  like  causes  will  always  be  connected  with  like 
effects,  every  question  in  proportion  must  give  one  of  the  following 
statements : 

1st  Cause     :     2d  Cause     =     1st  Effect     :     2d  Effect. 
1st  Effect     :     2d  Effect    =     1st  Cause     :     2d  Cause, 
in  which  the  two  causes  or  the  two  effects  forming  one  couplet, 
must  be  like  numbers  and  of  the  same  denomination. 

Considering  all  the  terms  of  a  proportion  as  abstract  numbers, 
we  may  say  that 

1st  Cause     :     1st  Effect     =     2d  Cause     :     2d  Effect. 
But  as  ratio  is  the  result  of  comparing  two  numbers  or  things 
of  the  same  kind,  (417),  the  first  form  is  regarded  as  the  more 
natural  and  philosophical. 

SIMPLE   PROPORTION. 

433.  Simple  Proportion  is  an  equality  of  two  simple  ratios, 
and  consists  of  four  terms. 

Questions  in  simple  proportion  involve  only  simple  causes  and 
simple  effects. 


250 


PROPORTION. 


FIRST    METHOD. 


1.  If  $S  will  buy  36  yards  of  velvet,  how  many  yards  may  be 


bought  for  $12? 


$  $ 

8      :      12 

1st  cause.    2d  cause. 


STATEMENT, 
yds. 

36 

1st  effect. 


yds. 


2d  effect 


8  X 


OPERATION. 

=     12  x  36 


ANALYSIS.  The  re- 
quired term  in  this  ex- 
ample is  an  effect ;  and 
the  statement  is,  $8  is 
to  $12  as  36  yards  is 
to  ( ?  ),  or  how  many 
yards.  Dividing  12  X 
3C,  the  product  of  the 
means,  by  8,  the  given 
extreme,  we  have  (  ?  ) 
=  54  yards,  the  re- 
quired term,  (428,11). 

2.  If  6  horses  will  draw  10  tons,  how  many  horses  will  draw 
15  tons? 

STATEMENT. 

tons. 

15 

2d  effect. 


54yd. 


horses.        horses.  tons. 

6     :     (?)     =     10 

1st  cause.    2d  cause.         1st  effect. 


OPERATION. 


ANALYSIS.  In  this  ex- 
ample a  cause  is  required  ; 
and  the  statement  is,  G 
horses  is  to  ( ? ),  or  how 
many  horses,  as  10  tons  ia 
to  15  tons.  Dividing  15  X 
6,  the  product  of  the  ex- 
tremes,  by  10,  the  given 

^ '  )        r  mean,  we  have  9,  the  re- 

(?)   =    9  horses.  quired  term,  (428,  I). 

434.    Hence  the  following 

RULE.  I.  Arrange  the  terms  in  the  statement  so  that  the  causes 
shall  compose  one  couplet,  and  the  effects  the  other,  putting  (?)  in 
the  place,  of  the  required  term. 

II.  If  the  required  term  be  an  extreme,  divide  the  product  of  the 
means  by  the  given  extreme;  if  the  required  term  be  a  mean, 
divide  the  product  of  the  extremes  by  the  given  mean. 

NOTES. — 1.  If  the  terms  of  nny  couplet  be  of  different  denominations,  they 
must  be  reduced  to  the  same  unit  value. 

2.  If  the   odd  term  be  a  compound  number,  it  must  be  reduced  either  to  its 
lowest  unit,  or  to  a  fraction  or  a  decimal  of  its  highest  unit. 

3.  If  the  divisor  and  dividend   contain  one  or  more  factors  common  to  both, 
they  should  be  canceled.     If  any  of  the  terms  of  a  proportion  contain  mixed 


SIMPLE  PROPORTION.  251 

numbers,  they  should  first  be  changed  to  improper  fractions,  or  the  fractional 
part  to  a  decimal. 

4.  When  the  vertical  line  is  used,  the  divisor  and  (?)  are  written  on  the  left, 
and  the  factors  of  the  dividend  on  the  right. 


SECOND    METHOD. 

The  following  method  of  solving  examples  in  simple 
proportion  without  making  the  statement  in  form,  may  be  used 
by  those  who  prefer  it. 

Every  question  in  simple  proportion  gives  three  terms  to  find  a 
fourth.  Of  the  three  given  terms,  two  will  always  be  like  numbers, 
forming  the  complete  ratio,  and  the  third  will  be  of  the  same  name  or 
kind  as  the  required  term,  and  may  be  regarded  as  the  antecedent  of 
the  incomplete  ratio ;  hence  the  required  term  may  be  found  by  mul- 
tiplying this  third  term,  or  antecedent,  by  the  ratio  of  the  other  two, 
(421,  II). 

From  the  conditions  of  the  question  we  can  readily  determine 
whether  the  answer,  or  required  term,  will  be  greater  or  less  than  the 
third  term ;  if  greater,  then  the  ratio  will  be  greater  than  1,  and  the 
two  like  numbers  must  be  arranged  in  the  form  of  an  improper  frac- 
tion, as  a  multiplier ;  if  less,  then  the  ratio  will  be  less  than  1,  and 
the  two  like  numbers  must  be  arranged  in  the  form  of  a  proper  frac- 
tion, as  a  multiplier. 

1.  If  4  tons  of  hay  cost  $36,  what  will  5  tons  cost? 

OPERATION.  ANALYSIS.      In  this  example,  4 

$36  X  -  =  $45,  Ans.          *ons  ^nd  5  tons  are  the  like  terms, 

and  $36  is  the  third  term,  and  of 

the  same  kind  as  the  answer  sought.  Now  if  4  tons  cost  $36,  will  5 
tons  cost  more,  or  less,  than  $36  ?  Evidently  more  :  and  the  required 
term  will  be  greater  than  the  third  term,  $36,  and  the  ratio  greater 
than  1.  We  therefore  arrange  the  like  terms  in  the  form  of  an  im- 
proper fraction,  {,  for  a  multiplier,  and  obtain  $45,  the  answer. 

2.  If  7  men  build  21  rods  of  wall  in  a  day,  how  many  rods  will 
4  men  build  in  the  same  time  ? 

OPERATION.  ANALYSIS.     In  this  example,  7 

21  ><  4  __  12  rods   Ans.          men  and  4  men  are  the  like  terms, 

and  21  rods  is  the  third  term,  and 

of  the  same  kind  as  the  answer  sought.  Since  4  men  will  perform  less 
work  than  7  men  in  the  same  time,  the  required  term  will  be  less  than 


252  PROPORTION. 

21,  and  the  ratio  less  than  1.  We  therefore  arrange  the  like  terms 
in  the  form  of  a  proper  fraction,  ^,  and  obtain  by  multiplication,  12 
rods,  the  answer. 

4-«SG.    Hence  the  following 

RULE.  I.  With  the  two  given  numbers,  ichich  are  of  the  same 
name  or  kind,  form,  a  ratio  greater  or  hss  than  1,  according  as  the 
answer  is  to  be  greater  or  less  than  the  third  given  number. 

II.  Multiply  the  third  number  by  this  ratio  j  the  product  will  be 
the  required  number  or  answer. 

NOTES. — 1.  Mixed  numbers  should  first  be  reduced  to  improper  fractions,  and 
the  ratio  of  the  fractions  found  according  to  418. 

2.  Reductions  and  cancellation  may  be  applied  as  in  the  first  method. 

The  following  examples  may  be  solved  by  either  of  the  fore- 
going methods. 

EXAMPLES    FOR   PRACTICE. 

1.  If  12  gallons  of  wine  cost  $30,  what  will  63  gallons  cost? 

2.  If  9  bushels  of  wheat  make  2  barrels  of  flour,  how  many 
barrels  of  flour  will  100  bushels  make  ?  A-ns.  22|. 

3.  If  18  bushels  of  wheat  be  bought  for  $22.25,  and  sold  for 
$26.75,  how  much  will  be  gained  on  240  bushels,  at  the  same  rate 
of  profit?  Ans.  $60. 

4.  If  6£  bushels  of  oats  cost  $3,  what  will  91  bushels  cost? 

5.  What  will  87.5  yards  of  cloth  cost,  if  If  yards  cost  $.42? 

6.  If  by  selling  $1500  worth  of  dry  goods  I  gain  $275.40,  what 
amount  must  I  sell  to  gain  $1000  ? 

7.  If  20  men  can  perform  a  piece  of  work  in  15  days,  how 
many  men  must  be  added  to  the  number,  that  the  work  may  be 
accomplished  in  4  of  the  time  ?  Ans.  5. 

8.  If  100  yd.  of  broadcloth  cost  $473.07^,  how  much  will 
3.25  yd.  cost? 

9.  If  1  Ib.  4  oz.  10  pwt,  of  gold  may  be  bought  for  $260.70, 
how  much  may  be  bought  for  $39.50  ?  Ans.  2  oz.  10  pwt. 

10.  In  what  time  can  a  man  pump  54  barrels  of  water,  if  he 
pump  24  barrels  in  1  h.  14  min.  ?        Ans.  2  h.  46  min.  30  sec. 

11.  If  |  of  a  bushel  of  peaches  cost  $^|,  what  part  of  a  bushel 
can  be  bought  for  $37g  ?  Ans.  T73  bu. 


COMPOUND   PROPORTION.  253 

12.  If  the  annual  rent  of  46  A.  134  P.  of  land  be  $374.70,  how 
much  will  be  the  rent  of  35  A.  90  P.  ? 

13.  If  a  man  gain  $1870.65  by  his  business  in  1  yr.  3  mo.,  how 
much  would  he  gain  in  2  yr.  8  mo.,  at  the  same  rate  ? 

14.  Two  numbers  are  to  each  other  as  5  to  7J,  and  the  less  is 
164.5,  what  is  the  greater?  Ans.  246. 7^. 

15.  If  16  head  of  cattle  require  12  A.  156  P.  of  pasture  during 
the  season,  how  many  acres  will  132  head  of  cattle  require? 

Ans.  107  A.  7  P. 

16.  If  a  speculator  in  grain  gain  $26.32  by  investing  $325,  how 
much  would  he  gain  by  investing  $2275? 

17.  What  will  be  the  cost  of  paving  an  open  court  GO. 5  ft.  long 
and  44  ft.  wide,  if  14.25  sq.  yd.  cost  $34i  ? 

18.  At  6J  cents  per  dozen,  what  will  be  the  cost  of  10  J  gross 
of  steel  pens  ? 

19.  If  when  wheat  is  7s.  6d.  per  bushel,  the  bakers'  loaf  will 
weigh  9  oz.;  what  ought  it  to  weigh  when  wheat  is  6s.  per  bushel? 

Ans.  Hi  oz. 


COMPOUND  PROPORTION. 

437.  Compound  Proportion  is  an  expression  of  equality  be< 
tween  a  compound  and  a  simple  ratio,  or  between  two  compound 
ratios. 

It  embraces  the  class  of  questions  in  which  the  causes,  or  the 
effects,  or  both,  are  compound.  The  required  term  must  be  either 
a  simple  cause  or  effect,  or  a  single  element  of  a  compound  cause 
or  effect. 

FIRST   METHOD. 

1.  If  8  men  mow  40  acres  of  grass  in  3  days,  how  many  acres 
will  9  men  mow  in  4  days  ? 

STATEMENT. 
1st  cause.        2d  cause.  1st  effect.          2d  effect 

=      40      :      (?) 
Or,  8  x  3  :  9  x  4      =      40      :      (?) 


I8-       J9 
(3-       J4 


254  PROPORTION. 

OPERATION.  ANALYSIS*      In  this  ex- 

,,js  9  X  4  X  40 ample  the    required   term 

"'  =          8x3  '  .        is  the  second  effect ;  and  the 

statement  is,  8  men  3  days 

is  to  9  men  4  days,  as  40  acres  is  to  (  ?  ),  or  how  many  acres.  Dividing 
the  continued  product  of  all  the  elements  of  the  means  by  the  ele- 
ments of  the  given  extreme,  we  obtain  (  ?  )  —  60  acres. 

2.  If  6  compositors  in  14  hours  can  set  36  pages  of  56  lines 
each,  how  many  compositors,  in  12  hours,  can  set  48  pages  of  54 
lines  each  ? 

STATEMENT. 

• 

1st  cadso.  2d  cause.        1st  effect.        2d  effect. 

(    6    .     {(?)..  f    36    .      r   48 
1 14         {    12    '  '  \    56          (54 

OPERATION.  ANALYSIS.     In  this  example,  an  element  of 

(?)      0  the   second  cause  is  required ;  and  the  state- 

/i?^  ment  is,  C  compositors  14  hours  is  to  (  ?)  com- 

7^$  positors  12  hours  as  36  pages  of  56  lines  each 

w  is  to  48  pages  of  54  lines  each.    Now,  since  the 

/  \  __  9^  Ans.  required  term  is  an  element  of  one  of  the  means, 
we  divide  the  continued  product  of  all  the  ele- 
ments of  the  extremes  by  the  continued  product  of  all  the  given  ele- 
ments of  the  means.  Placing  the  dividend  on  the  right  of  the  verti- 
cal line  and  the  divisors  on  the  left,  and  canceling  equal  factors  wo 
obtain  (  ?  )  =  9. 

438.   From  these  illustrations  we  deduce  the  following 
RULE.     J.    Of  the  given  terms,  select  those  which  constitute  the 
causes,  and  those  which  constitute  the  effects,  and  arrange  them  in 
couplets,  putting  (?)  in  place  of  the  required  term. 

II.  Then,  if  the  blank  term  (?)  occur  in  either  of  the  extremes, 
divide  the  product  of  the  means  by  the  product  of  the  extremes; 
but  if  the  blank  term  occur  in  either  mean,  divide  the  product  of 
the  extremes  by  the  product  of  the  means. 

NOTKS. 1.  The  onuses  must  he  exactly  alike  in  the  number  and  "kind  of  their 

term?  :  the  same  is  true  of  the  effects. 

2.  The  snme  preparation  of  the  terms  by  reduction  is  to  be  observed  as  in 
friuiple  proportion. 


COMPOUND  PROPORTION,  255 

SECOND    METHOD. 

The  second  method  given  in  Simple  Proportion,  is  also 
applicable  in  Compound  Proportion. 

In  every  example  in  compound  proportion  all  the  terms  appear  in 
couplets,  except  one,  called  the  odd  term,  which  is  always  of  the  same 
kind  as  the  answer  sought.  Hence  the  required  term  in  a  compound 
proportion  may  be  found,  by  multiplying  the  cdd  term  by  the  com- 
pound ratio  composed  of  all  the  simple  ratios  formed  by  these  couplets, 
each  couplet  being  arranged  in  the  form  of  a  fraction. 

The  fraction  formed  by  any  couplet  will  be  improper  when  the  re- 
quired term,  considered  as  depending  on  this  couplet  alone,  should 
be  greater  than  the  odd  term ;  and  proper,  when  the  required  term 
should  be  less  than  the  odd  term. 

1.  If  it  cost  $4320  to  supply  a  garrison  of  32  men  with  pro- 
visions for  18  days,  when  the  rations  are  15  ounces  per  day,  what 
will  it  cost  to  supply  a  garrison  of  24  men  34  days,  wn*en  the 
rations  are  12  ounces  per  day  ? 

OPERATION, 
men.          days.        ounces. 

$4320   x  !£   X   ||   X   ||  =  $4896 

ANALYSIS.  In  this  example  there  are 
three  pairs  of  terms,  or  couplets,  viz.,  32 
men  and  24  men,  18  days  and  34  days,  15 
ounces  and  12  ounces ;  and  there  is  an  odd 
term,  $4320,  which  is  of  the  same  kind  as 
=  $4896,  Ans.  the  required  term.  We  arrange  each  coup- 
let as  a  multiplier  of  this  term,  thus; 
First,  if  it  cost  $4320  to  supply  32  men,  will  it  cost  more,  or  less,  to 
supply  24  men  ?  Less  ;  we  therefore  arrange  the  couplet  in  the  form  of 
a  proper  fraction  as  a  multiplier,  and  we  have  $4320  X  j|.  Next,  if  it 
cost  $4320  to  supply  a  garrison  18  days,  will  it  cost  more,  or  less,  to 
supply  it  34  days?  More  ;  hence  the  multiplier  is  the  improper  frac- 
w'on  }|,  and  we  have  $4320  X  $4  X  ^.  Next,  if  it  cost  $4320  to 
supply  a  garrison  with  rations  of  15  ounces,  will  it  cost  more,  or  less, 
when  the  rations  are  12  ounces  ?  Less ;  consequently,  the  multiplier 
is  the  proper  fraction  {^  and  we  have  $4320  X  £4  X  f  £  X  }£  =$4896, 
the  required  term.  Hence  the  following 


256  PROPORTION. 

RULE.  I.  Of  the  terms  composing  each  couplet  form  a  ratio 
greater  or  less  than  1,  in  the  same  manner  as  if 'the  answer  de- 
pended on  those  two  and  the  third  or  odd,  term. 

II.  Multiply  together  the  third  or  odd  term  and  these  ratios; 
the  product  will  be  the  answer  souyht. 

EXAMPLES    FOR   PRACTICE. 

1.  If  12  horses  plow  11   acres  in  5  days,  how  many  horses 
would  plow  33  acres  in  18  days?  Ans.  10. 

2.  If  480  bushels  of  oats  will  last  24  horses  40  days,  how  long 
will  300  bushels  last  48  horses,  at  the  same  rate  ? 

Ans.  12  J  days. 

3.  If  7  reaping  machines  can  cut  1260  acres  in  12  days,  in 
how  many  days  can  16  machines  reap  4728  acres  ? 

Ans.  19.7  days. 

4.  If  144  men  in  6  days  of  12  hours  each,  build  a  wall  200  ft. 
long,  3  ft.  high,  and  2  ft.  thick,  in  how  many  days  of  7  hours 
each  can  30  men  build  a  wall  350  ft.  long,  6  ft.  high,  and  3  ft. 
thick?  Ans.   259.2  da. 

5.  In  how  many  days  will  6  persons  consume  5  bu.  of  potatoes, 
if  3  bu.  3  pk.  last  9  persons  22  days  ? 

6.  How  many  planks  lOf  ft.  long  and  li  in.  thick,  are  equiva- 
lent to  3000  planks  12  ft.  8  in.  long  and  2|  in.  thick? 

Ans.  6531 J. 

7.  If  300  bushels  of  wheat  @  $1.25  will  discharge  a  certain 
debt,  how  many  bushels  @  $.90  will  discharge  a  debt  3  times  as 
great?  Am.  1250  bu. 

8.  If  468  bricks,  8  inches  long  and  4  inches  wide,  arc  required 
for  a  walk  26  ft.  long  and  4  ft.  wide,  how  many  bricks  will  be 
required  for  a  walk  120  ft.  long  and  6  ft.  wide  ? 

9.  If  a  cistern  17  J  ft.  long,  10  J  ft.  wide,  and  13  ft.  deep,  hold 
546  barrels,  how  many  barrels  will  a  cistern  hold  that  is  16  ft. 
long,  7  ft.  wide,  and  15  ft.  deep?  Ans.  384  bbl. 

10.  If  11  men  can  cut  147  cords  of  wood  in  7  days,  when  they 
work  14  hours  per  day,  how  many  days  will  it  take  5  men  to  cut 
150  cords,  working  10  hours  each  day? 


PROMIPCFOUS  EXAMPLES  257 

PROMISCUOUS    EXAMPLES    IN    PROPORTION. 

1.  If  a  staff  4  ft.  long  cast  a  shadow  7  ft.  in  length,  what  is 
the  liiglit  of  a  tower  that  casts  a  shadow  of  198  ft.  at  the  same 
time?  Am.  113^  ft. 

2.  A  person  failing  in  business  owes  $972,  and  his  entire  prop- 
erty is  worth  but  $607.50;  how  much  will  a  creditor  receive  on  a 
debt  of  $11.33£  ?  Ans.  S7.08+. 

3.  If  3  cwt.  can  be  carried  660  mi.  for  $4,  how  many  cwt.  can 
be  carried  60  mi.  ilr  $12  ?  Ans.  99. 

4.  A  man  can  perform  a  certain  piece  of  work  in  18  days  by 
working  8  hours  a  day ;  in  how  many  days  can  he  do  the  same 
work  by  working  10  hours  a  day  ?  -4ns.  14|. 

5.  How  much  land  worth  $16.50  an  acre,  should  be  giv  n  in 
exchange  for  140  acres,  worth  $24.75  an  acre? 

6.  If  I  gain  $155.52  on  $1728  in  1  yr.  6  mo.,  how  much  will 
I  gain  on  $750  in  4  yr.  6  mo.?  Ans.  $202.50. 

7.  If  1  lb.  12  oz.  of  wool  make  2}  yd.  of  cloth  6  qr.  wide,  how 
many  lb.  of  wool  will  it  take  for  150  yd.  of  cloth  4  qr.  wide  ? 

8.  What  number  of  men  must  be  employed  to  finish  a  piece  of 
work  in  5  days,  which  15  men  could  do  in  20  days?     An*.  60. 

9.  At  12s.  7d.  per  oz.,  N.  Y.  currency,  what  will  be  the  cost 
of  a  service,  of  silver  plate  weighing  15  lb.  11  oz.  13  pwt.  17  gr.  ? 

10.  If  a  cistern  16  ft.  long,  7  ft.  wide,  and  15  ft.  deep,  cost 
$36.72,  how  much,  at  the  same  rate  per  cubic  foot,  would  another 
cistern  cost  that  is  17£  ft.  long,  10£  ft.  wide,  and  16  ft.  deep? 

11.  A  borrows  $1200   and   keeps  it  2  yr.   5  mo.  5  da.;  what 
sum  should  he  lend  for  1  yr.  8  mo.  to  balance  the  favor  ? 

12.  A  farmer  has  hay  worth  $9  a  ton,  and  a  merchant  has  flour 
worth  $5  per  barrel.     If  in  trading  the  former  asks  $10.50  for 
his  hay,  how  much  should  the  merchant  ask  for  his  flour  ? 

13.  If  12  men,  working  9  hours  a  day  for  15|  days,  were  able 
to  execute  f  of  a  job,  how  many  men  may  be  withdrawn  and  the 
job  be  finished  in  15  days  more,  if  the  laborers  are   employed 
only  7  hours  a  day  ?  -4ns.  4. 

14.  If  the  use  of  $300  for  1  yr.  8  mo.  is  worth  $30,  how  much 
is  the  use  of  $210.25  for  3  yr.  4  mo.  24  da.  worth  ? 

22*  a 


258  PROPORTION. 

15.  What  quantity  of  lining  f  yd.  wide,  will  it  require  to  line 
9£  yd.  of  cloth,  H  yd.  wide?  A?is.   15|  yd.. 

16.  If  it  cost  $95.60  to  carpet  a  room  24  ft.  by  18  ft.,  how 
much  will  it  cost  to  carpet  a  room  38  ft.  by  22  ft.  with  the  same 
material?  Ans.  $185.00+. 

17.  If  16/H  cords  of  wood  last  as  long  as   11,°,  tons  of  coal, 
how  many  cords  of  wood  will  last  as  long  as  15T73  tons  of  coal? 

18.  A  miller  has  a  bin  8  ft.  long,  4i  ft.  wide,  and  2^  ft.  deep, 
and  its  capacity  is  75  bu.  ;  how  deep  must  he  make  another  bin 
which  is  to  be  18  ft.  long  and  3|  feet  wide,  that  its  capacity  may 
be  450  bu.  ?  Ans.  7^  ft. 

19.  If  4  men  in  2J  days,  mow  Gf  acres  of  grass,  by  working 
8£  hours  a  day,  how  many  acres  will  15  men  mow  in  3£  days,  by 
working  9  hours  a  day  ?  Ans.  40  jfi  acres. 

20.  If  an  army  of  600  men  have  provisions  for  5  weeks,  allowing 
each  man  12  oz.  a  day,  how  many  men  may  be  maintained  10 
weeks  with  the  same  provisions,  allowing  each  man  8  oz.  a  day  ? 

21.  A  cistern  holding  20  barrels  has  two  pipes,  by  one  of  which 
it  receives  120  gallons  in  an  hour,  and  by  the  other  discharges 
80  gallons  in  the  same  time;  in  how  many  hours  will  it  be  filled? 

22.  A  merchant  in  selling  groceries  sells  14^  oz.  for  a  pound; 
how  much  does  he  cheat  a  customer  who  buys  of  him  to  the  amount 
of  $38.40  ?  Ans.  $3.45. 

23.  If  5  Ib.  of  sugar  costs  $.62£,  and  8  Ib.  of  sugar  are  worth 
5  Ib.  of  coffee,  how  much  will  75  Ib.  of  coffee  cost  ? 

24.  B  and  C  have  each  a  farm;  B's  farm  is  worth  $32.50  an 
acre,  and  C's  $28.75 ;  but  in  trading  B  values  his  at  $40  an  acre. 
What  value  should  C  put  upon  his  ? 

25.  If  it  require  $59f  reams  of  paper  to  print  12000  copies  of 
an  8vo.  book  containing  550  pages,  how  many  reams  will  be  required, 
to  print  3000  copies  of  a  12mo.  book  containing  320  pages? 

26.  If  248  men,  in  5J  days  of  12  hours  each,  dig  a  ditch  of  7 
degrees  of  hardness,  23'2i  yd.  long,  S.t  yd.  wide,  and  2£  yd.  deep; 
in  how  many  days  of  9  hours  each,  will  24  men  dig  a  ditch  of  4 
degrees  of  hardness,  387  £  yd.  long,  5^  yd.  wide,  and  3*  yd  deep  ? 

Ans.  155 


NOTATION.  259 


PERCENTAGE. 

440.  Per   Cent,    is  a  contraction   of  the   Latin  phrase  per 
centum,  and  signifies  I?/  the  hundred  ;  that  is,  a  certain  part  of 
every  hundred,  of  any  denomination  whatever.     Thus,  4  per  cent 
means  4  of  every  hundred,  and  may  signify  4  cents  of  every  100 
cents,  4  dollars  of  every  100  dollars,  4  pounds  of  every  100 
pounds,  etc. 

NOTATION. 

441.  The  character,   %,  is  generally  employed  in  business 
transactions  to  represent  the  words  per  cent. ;   thus  C  %  signifies 
6  per  cent. 

44^.  Since  any  per  cent,  is  some  number  of  hundredths,  it  is 
properly  expressed  by  a  decimal  fraction;  thus  5  per  cent. 
s=s  5  %  =  .05.  Per  cent,  may  always  be  expressed,  however, 
either  by  a  decimal  or  a  common  fraction,  as  shown  in  the  following 

TABLE. 


"Words. 

1    per  cent. 
2    per  cent. 
4    per  cent. 
5    per  cent. 
6    per  cent. 
7    per  cent. 
8    per  cent. 
10    per  cent. 
20    per  cent. 
25    per  cent. 

Symbols.           Decimals. 

=        1    %     =      .01 
=        2    %     =      .02 
=        4    %     =      .04 
=        5    %     -       .05 
=        6    %     =       .06 
=        7    %     =      .07 
=        8    %     =       .08 
=      10    %     =      .10 
=      20    %     =      .20 
=      25    %     =      .25 

Common  tractions. 

=         Toff         =      TOff 
—        Tffff         ==       s'ff 

j               _y 

5                        jl 
Tffff                      10 

—         Iff*         ==      T»3 
••         T¥f         *"        A 

=               S!t              1 
100                           4 

50    per  cent. 

=      50    %     = 

.50 

5  0 

=          ^ 

100    per  cent. 

=     100    %     = 

1.00 

=       ins 

=          1 

125    per  cent. 

—     125     %     = 

1.25 

~         T    ff 

5 

( 

\  per  cent. 

=               3    2      = 

.001 

=             Too 

==        259 

I  per  cent. 

I    %      = 

.00^ 

—             4 
flf 

=    «£* 

12^  per  cent 

=       12J  *     = 

.12£ 

—             '2^ 

100 

=       i 

260  PERCENTAGE. 

EXAMPLES    FOR   PRACTICE. 

»• 

1.  Express  decimally  3  per  cent.  ;  9  per  cent.  ;  12  per  cent.  ; 
16  per  cent.;  23  per  cent.  ;  37  per  cent.;  75  per  cent.;  125  per 
cent.  ;  184  per  cent.  ;  205  per  cent. 

2.  Express  decimally  15  %  ;  11  %  ;  4*  %  ;  5^  %  ;  8£  %  ', 
20*  %  ;  25|  %  ;  35f  %  ;  241  %  ',  130*  %. 

3.  Express  decimally  \  per  cent.  ;  f  per  cent.  ;  £  per  cent.  ; 
|  per  cent.;  f  per  cent.;  275  per  cent.;  -ff^  per  cent.;  lT5g  per 
cent.  ;  lOi  per  cent. 

4.  Express  by  common  fractions,  in  their  lowest  terms,  4  %  ; 
37*  %  ;  16i  %  ;  11>  %  ;  42?  %  ;  45ft  %  ;  48j>T  #. 

5.  What  per  cent,  is  .0725  ? 

ANALYSIS.    .0725  =  .07£  =  7J  %,  Ans. 

6.  What  per  cent,  is  .065?  Ans.  6i  %. 

7.  What  per  cent,  is  .14375?  Ans.  14f  %. 

8.  What  per  cent,  is  .0975  ? 

9.  What  per  cent,  is  .014  ? 

10.  What  per  cent,  is  .1025? 

11.  What  per  cent,  is  .004  ? 

12.  What  per  cent,  is   028  ? 

13.  What  %  is  .1324? 

14.  What  %  is  .084f  ? 

15.  What  %  is  .004  Jy  ?  Ans.  T«T  % 


16.  What  %  is  .003^  ? 


GENERAL  PROBLEMS  IN  PERCENTAGE. 

In  the  operations  of  Percentage  there  are  five  parts  01 
elements,  namely  :  Rate  per  cent.,  Percentage,  Base,  Amount,  and 
Difference. 

444.  Hate  per  Cent,,  or  Rate,  is  the  decimal  which  denotes 
how  many  hundredths  of  a  number  are  to  be  taken. 

NOTKS. — 1.  Suoh  expressions  as  6  per  cent.,  nnd  5  %,nre  esscntinlly  dn-iwah, 
{he  words  p^.r  ct-ft.,  or  the  clmmcter  %,  indicating  the  deciinnl  deiioiiiiiiiitor. 

2.  If  the  di'i-iinal  he  re«luced  to  !i  common  fraction  in  its  lowtxt  terms,  this 
fraction  will  still  be  the  equivalent  rate,  though  not  the  rate  per  cent. 


PROBLEMS  IN  PERCENTAGE.  261 

445.  Percentage  is  that  part  of  any  number  which  is  indi- 
cated by  the  rate. 

446.  The  Base  is  the  number  on  which  the  percentage  is 
computed. 

447.  The  Amount  is  the  sum  obtained  by  adding  the  per- 
centage to  the  base. 

448.  The  Difference  is  the  remainder  obtained  by  subtract- 
ing the  percentage  from  the  base. 

PROBLEM    I. 

449.  Given,   the   base   and   rate,  to   find   the   per- 
centage. 

1.  What  is  5  %  of  360? 

ANALYSIS.     Since   5    fo    of    any 
OPERATION.  number   is    .05    of    that    number, 


(442),  we  multiply  the  base,  3GO, 

.'._!  by   the   rate,  .05,  and   obtain   the 

18.00,  Ans.  percentage,  18.     Or,  since  the  rate 

Or,  is  T£(T  •=  2>ff,  we  have  3GO  X  Jff  == 

360  X  TT'JJ  =18,  Ans.  18»  tne  percentage.    Hence  the  fol- 

lowing 
RULE.      Multiply  the  base  l>y  the  rate. 

NOTE  1. — Percentage  is  always  a  product,  of  which  the  base  nnd  rate  are  the 
fuctors. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  4  per  cent,  of  250  ?  Ans.  10. 

2  What  is  7  per  cent,  of  3500  ?  Ans.  245. 

3  What  is  16  per  cent,  of  324  ?  Ans.  51.84. 

4  What  is  12£  per  cent,  of  $5600?  Ana.  -S700. 

5.  What  is  9  %  of  785  Ibs.? 

6.  What  is  25  %  of  960  mi.  ? 

7.  What  is  75  %  of  487  bu.?  Ans.  365.25  bu. 

8.  What  is  33^  %  of  2757  men? 

9.  What  is  125  %  of  756? 

10.  What  is  |  %  of  $2364  ?  Am.  $5.91. 


262  PERCENTAGE. 

11.  What  is  3§  %  of  $856?  Arts.  $31,39  -. 

12.  What  is  |  %  Of  5?  Amt  _i^ 

13.  What  is  14*    %  of  51  ? 

14.  If  the  base  is  $375,  and  the  rate  .05,  what  is  the  percent- 
age ?  A tis.  $18.75. 

15.  A  man  owed  $536  to  A,  $450  to  B,  and  $784  to  C;  how 
much  money  will  be  required  to  pay  54  %  of  his  debts  ? 

16.  My  salary  is  $1500  a  year;  if  I  pay  15  %  for  board,  5  % 
for  clothing,  6  %  for  books,  and  8  %   for  incidentals,  what  are 
my  yearly  expenses  ?  Ans.  $510. 

NOTK  2.  — 15  %  +  5  %  +  6  %  +  8  %  =  34  %.  In  all  cases  where  several 
rates  refer  to  the  same  base,  they  may  be  added  or  subtracted,  according  to  the 
conditions  of  the  question. 

17.  A  man  having  a  yearly  income  of  $3500,  spends  10  per 
cent  of  it  the  first  year,  12  per  cent,  the  second  year,  and  18  per 
cent,  the  third  year;  how  much  does  he  save  in  the  3  years? 

18.  A  had  $6000  in  a  bank.     He  drew  out  25  %  of  it,  then 
30  %  of  the  remainder,  and  afterward  deposited  10  %  of  what 
he  had  drawn  ;  how  much  had  he  then  in  bank  ?    Ans.  $3435. 

19.  A  merchant  commenced  business,  Jan.  1,  with  a  capital  of 
$5400,  and  at  the  end  of  1  year  his  ledger  showed  the  condition 
of  his  business  as  follows :  For  Jan.,  2  %  gain  ;  Feb.,  3^  %  gain  ; 
March,  \  ff0  loss;   Apr.,  2  %  gain;   May,  2£  %  gain;   June,  If 
%  loss;    July,  U  %  gain;   Aug.,  1  %  loss;    Sept.,  2|  %  gain; 
Oct.,  4  %  gain>   Nov.,  f  %  loss;   Dec.,  3  %  gain.      What  were 
the  net  profits  of  his  business  for  the  year?  Ans.  $918. 

PROBLEM    II. 

45O.  Given,  the  percentage  and  base,  to  find  the 
rate- 

1.  What  per  cent  of  360  is  18  ? 

OPERATION.  ANALYSIS.    Since  the  percent- 

18  -i-  360  =  .05  =  5  ffo  aSe  ^s  always  the  product  of  the 

Qr  base  and  rate,  (449),  we  divide 

is          i          05  =  5  tf  the  glvcn  PercentaSe>  18'  hj  the 

given  base,  360,  and  obtain  the 

r.equired  rate,  .05  =  5  % .     Hence  the 


PROBLEMS  IN  PERCENTAGE.  2G3 

• 

RULE.      Divide  the  percentage  by  the  base. 

EXAMPLES    FOR    PRACTICE. 

1.  What  per  cent,  of  $720  is  $21.00  ?  Ans.  3. 

2.  What  per  cent,  of  1500  Ib.  is  234  Ib.  ? 

3.  What  per  cent,  of  980  rd.  is  49  rd.  ? 

4.  What  per  cent,  of  £320  10s.  is  £25  12.8s.?  Ans.  8. 

5.  What  per  cent,  of  46  gal.  is  5  gal.  3  qt.?  Ans.   12  J. 
G.  What  per  cent,  of  7.85  mi.  is  5.495  mi.?  Ans.  70, 

7.  What  per  cent,  of  T85  is  |  ?  Ans.  75. 

8.  What  per  cent,  of  4  is  ^  ? 

9.  What  per  cent,  of  560  is  80  ? 

10.  The  base  is  $578,  and  the  percentage  is  $26.01 ;  what  is  the 
rate?  Ans.  4£  <J0. 

11.  The  base  is  $972.24,  and  the  percentage  is  $145.836;  what 
is  the  rate  ? 

12.  An  editor  having  5600  subscribers,  lost  448;  what  was  his 
loss  per  cent  ?  Ans.   8. 

13.  A  merchant  owes  $7560,  and  his  assets  are  $4914;  what 
per  cent,  of  his  debts  can  he  pay  ?  Ans.   65. 

14.  A  man  shipped  2600  bushels  of  grain  from  Chicago,  and 
455  bushels  were  thrown  overboard  during  a  gale ;  what  was  the 
rate  per  cent,  of  his  loss  ? 

15.  A  miller  having  720  barrels  of  flour,  sold  288  barrels;  what 
per  cent,  of  his  stock  remained  unsold  ?  Ans.  60. 

16.  What  per  cent,  of  a  number  is  30  %  of  f  of  it  ? 

17.  The  total  expenditures  of  the  General  Government,  for  the 
year  ending  June  30,  1858,  were  $83,751,511.57;  the  expenses 
of  the  War  Department  were  $23,243,822.38,  and  of  the  Navy 
Department,  $14,712,610.21.     What  per  cent,  of  the  whole  ex- 
pense of  government  went  for  armed  protection  ? 

Ans.  45£,  nearly. 

18.  In  the  examination  of  a  class,   165  questions  were  sub- 
mitted to  each  of  the  5  members ;  A  answered  130  of  them,  B 
125,  C  96,  D  110,  and  E  160.     What  was  the  standing  of  the 
class?  Ans.  75.27  %. 


264  PERCENTAGE. 

PROBLEM    ITT. 

451.   Given,  the   percentage   and   rate,  to  find  the 
base. 

I.   18  is  5  %  of  what  number? 

OPERATION.  ANALYSIS.     Since   the  percent- 

•IQ         n-        o™     A  aSe  is  always  the  product  of  the 

30'  Ans'  base  and  rate,  (449),  we  divide 

Or,  the  given  percentage,  18,  by  the 

18  -^  ^  =  360,  Ans.  given  rate,  .05,  or  Jff,  and  obtain 

the  base,  360.     Hence  the 
RULE.     Divide  the  percentage  ly  the  rate. 


EXAMPLES    FOR   PRACTICE. 

1.  18  is  25  %  of  what  number?  Ans.  72. 

2.  54  is  15  %  of  what  number? 

3.  17.5  is  2£  %  of  what  number?  Ans.  750. 

4.  2.28  is  5  %  of  what  number? 

5.  414  is  120  %  of  what  number? 

6    6119  is  105J  %  of  what  number?  Ans.   5800. 

7.  .43  is  711  %  of  what  number?  Ans.  .6. 

8.  The  percentage  is  $18.75,  and  the  rate  is  2  J  %  ;  what  is  the 
base  ?  Ans.  $750. 

9.  The  percentage  is  31J,  and  the  rate  31J  %  ;  what  is  the 
base? 

10.  I  sold  my  house  for  $4578,  which  was  84  %  of  its  cost; 
what  was  the  cost  ?  Ans.  $5450. 

11.  A  wool  grower  sold  3150  head  of  sheep,  and  had  80  %  of 
his  original  flock  leffcj  how  many  sheep  had  he  at  first? 

12.  A  man  drew  40  %  of  his  bank  deposits,  and  expended  13| 
<fo  of  the  money  thus  drawn  in  the  purchase  of  a  carriage  worth 
$116;  how  much  money  had  he  in  bank?  Ans.  $2175. 

13.  If  $147.56   is   13|    %  of  A's   money,  and  4J   %  of  A's 
money  is  8  %  of  B's,  how  much  more  money  has  A  than  B  ? 

AM.  $461.12i 


PROBLEMS  IN  PERCENTAGE. 

14.  In  a  battle  4  %  of  the  army  were  slain  upon  the  field;  and  5 
ffo  of  the  remainder  died  of  wounds,  in  the  hospital.     The  differ- 
ence between  the  killed  and  the  mortally  wounded  was  168 ;  how 
many  men  were  there  in  the  army  ?  Ans.  21000. 

NOTK.— 100  %  —  4  %  =  96  %,  left  after  the  battle;  and  5  %   of  96  %  = 
4*  %,  the  part  of  the  artny  that  died  of  wounds. 

15.  A  owns  f  of  a  prize  and  B  the  remainder;  after  A  has 
taken  40  %  of  his  share,  arid  B  20  %  of  his  share,  the  remainder 
is  equitably  divided  between  them  by  giving  A  $1950  more  than 
B ;  what  is  the  value  of  the  prize  ?  Ans.  $7800. 

PROBLEM    IV. 

452.  Given,  the  amount  and  rate,  to  find  the  base. 
1.  What  number  increased  by  5  %  of  itself  is  equal  to  £78  ? 

OPERATION.  ANALYSIS.      If   any  number 

1  -f-     .05  =  1.05  ke  increased  by  5  %  of  itself 

378  -r-  1.05  =  360,  Ans.  the  amount  will  be  1.05  times 

the  number.     We  therefore  di- 
Or,  vide  the  given  amount,  378,  by 

1  -\-  ^  =    |^  1.05,  or  |£,  and  obtain  the  base, 

378  -+-  2  J  _-  360,  Ans.  360,  which  is  the  number  re- 

quired.    Hence  the 

RULE.     Divide  the  amount  ly  1  plus  the  rate. 

NOTE  1. — The  amount  is  always  a  product,  of  which  the  base  is  one  factor, 
and  1  plus  the  rate  the  other  factor. 

EXAMPLES    FOR   PRACTICE. 

1.  What  number  increased  by  15  %  of  itself  is  equal  to  644  ? 

Ans.  560. 

2.  A  has  $815.36,  which  is  4  %  more  than  B  has;  how  much 
money  has  B  ?  Ans.  $784. 

3.  Having  increased  my  stock  in  trade  by  12   %  of  itself,  I 
find  that  I  have  13800 ;  how  much  had  I  at  first  ? 

4.  Ic  1860  the  population  of  a  certain  city  was  39600,  which 
was  an  increase  of  1(1  %  during  the  10  years  preceding;  what 
was  the  population  in  1850  ? 


266  PERCENTAGE. 

5.  My  crop  of  wheat  this  year  is  8  %  greater  than  my  crop  of 
last  year,  and  I  have  raised  during  the  two  years  5200  bushels; 
what  was  my  last  year's  crop  ?  Ans.  2500  bu. 

NOTE  2.  — 1.00  +  1.08  =  2.08.     Hence,  5200  bu.  =  2.08  %  of  last  year's  crop. 

6.  The  net  profits  of  a  nursery  in  two  years  were  $6970,  and 
the  profits  the  second  year  were  5  %  greater  than  the'  profits  the 
first  year ;  what  were  the  profits  each  year  ? 

Ans.   1st  year,  $3400 ;  2d  year,  $3570. 

7.  If  a  number  be  increased  8  %,  and  the  amount  be  increased 
7  tfoi  tne  result  will  be  86.67  ;  required  the  number 

NOTE  3. —  The  whole  amount  will  be  1.08  X  1.07  =  1.155C  times  the  original 
number. 

8.  A  produce  dealer  bought  grain  by  measure,  and  sold  it  by 
weight,  thereby  gaining  1J  %   in  the  number  of  bushels.     He 
sold  at  a  price  5  %  above  his  buying  price,  and  received  34910.976 
for  the  grain  •  required  the  cost.  Ans.  $4608. 

9.  B  has  6  %,  and  04%  more  money  than  A,  and  they  all 
have  $11160  ;  how  much  money  has  A  ?  Ans.  $3600. 

10.  In  the  erection  of  a  house  I  paid  twice  as  much  for  mate- 
rial as  for  labor.    Had  I  paid  6  %  more  for  material,  and  9  %  more 
ior  labor,  my  house  would  have  cost  $1284 ;  what  was  its  cost  ? 

Ans.  $1200. 

PROBLEM   V. 

453.   Given,  the  difference  and  rate,  to  find  the  base. 
1.  What  number  diminished  by  5  %  of  itself,  is  equal  to  342  ? 

OPERATION.  ANALYSIS.     If  any  number  be  di- 

1 _Q5  _   95  minished  by  5  %  of  itself,  the  dif- 

342  -s-  .95  =  360,  Ans.  ference  will  be  .95  of  the  number. 

Or  We  therefore  divide  the  given  differ- 

1 ^   '—  JMJ  ence,  342,  by  .95,  or  £g,  and  obtain 

342  -f-  12   =  360,  Ans.  the  base,  3GO,  which  is  the  required 
number.     Hence  the 

RULE.      Divide  the  difference  by  1  minus  the  rate. 

NOTB. — The  difference  is  always  a  product,  of  which  the  base  is  one  factor, 
and  1  minus  the  rate  the  other. 


PROBLEMS  IN  PERCENTAGE.  267 

EXAMPLES    FOR    PRACTICE. 

1.  What  number  diminished  by  10  %  of  itself  is  equal  to  504? 

Ans.   560. 

2.  The  rate  is   8   <fr,  and  the  difference  $4.37;  what  is  the 
base? 

3.  After  taking  away  15  %  of  a  heap  of  grain,  there  remained 
40  bu.  3£  pk. ;  how  many  bushels  were  there  at  first? 

Ans.  48  bu. 

4.  Having  sold  36  %  of  my  land,  I  have  224  acres  left;  how 
much  land  had  I  at  first  ? 

5.  After  paying  65  %  of  my  debts,  I  find  that  $2590  will  dis- 
charge the  remainder;  how  much  did  I  owe  in  all? 

Ans.  $7400. 

6.  A  young  man  having  received  a  fortune,  deposited  80  o/0  of 
it  in  a  bank.     He  afterward  drew  20  %  of  his  deposit,  and  then 
had  $5760  in  bank ;  what  was  his  entire  fortune  ? 

Ans.  $9000. 

7.  A  man  owning  f-  of  a  ship,  sold  12  ffc  of  his  share  to  A,  and 
the  remainder  to  B,  at  the  same  rate,  for  $20020;  what  was  the 
estimated  value  of  the  whole  ship  ?  Ans.  $26000. 

8.  An  army  which  has  been  twice  decimated  in  battle,  now 
contains  only  6480  men ;  what  was  the  original  number  in  the 
army?  Ans.   8000. 

v  9.  Each  of  two  men,  A  and  B,  desired  to  sell  his  horse  to  C. 
A  asked  a  certain  price,  and  B  asked  50  %  more.  A  then  re- 
duced his  price  20  '%,  and  B  his  price  30  %.  at  which  prices  C 
took  both  horses,  paying  for  them  $148 ;  what  was  each  man's 
asking  price  ?  4  j  A,  $80. 

(B,  $120. 

10.  A  buyer  expended  equal  sums  of  money  in  the  purchase  of 
wheat,  corn,  and  oats.  In  the  sales,  he  cleared  6  %  on  the  wheat, 
and  3  %  on  the  corn,  but  lost  17  %  on  the  oats;  the  whole 
amount  received  was  $2336.  What  sum  did  he  lay  out  in  each 
kind  of  grain  ?  Ans.  $800. 


268  PERCENTAGE. 

APPLICATIONS  OF  PERCENTAGE. 

t 

454.  The  principal  applications  of  Percentage,  where  time  is 
not  considered,  are  Commission,  Stocks,  Profit  and  Loss,  Insurance, 
Taxes,  and  Duties.     And  since  the  five  problems  in  Percentage 
involve  all  the  essential  relations  of  the  parts  or  elements,  we  have 
for  the  above  applications  the  following 

GENERAL  RULE.  Note  what  elements  of  Percentage  are  given 
in  the  example,  and  what  element  is  required  ;  then  apply  the  spe- 
cial rule  for  the  corresponding  case. 

COMMISSION. 

455.  An  Agent,  Factor,  or  Broker,  is  a  person  who  trans- 
acts business  for  another. 

456.  A  Commission  Merchant  is  an  agent  who  buys  and 
sells  goods  for  another. 

45  T.  Commission  is  the  fee  or  compensation  of  an  agent, 
factor,  or  commission  merchant. 

458.  A  Consignment  is  a  quantity  of  goods  sent  to  one  person 
to  be  sold  on  commission  for  another  person. 

450.  A  Consignee  is  a  person  who  receives  goods  to  sell  for 
another;  and 

46®.  A  Consignor  is  a  person  who  sends  goods  to  another  to 
be  sold. 

461.  The  Net  Proceeds  of  a  sale  or  collection  is  the  sum 
left,  after  deducting  the  commission  and  other  charges. 


.  —  A  person  who  is  employed  in  establishing  mercantile  relations  between 
others  living  at  a  distance  from  each  other,  is  called  the  Corre«p<mdent  of  the 
party  in  whose  behalf  he  acts.  A  correspondent  is  the  agent  of  those  whose 
custom  or  patronage  he  secures  to  the  party  in  whose  interest  he  is  employed. 

46^.  Commission  is  usually  reckoned  at  a  certain  per  cent,  of 
the  money  involved  in  the  transaction  ;  hence  we  have  the  follow- 
ing relations  : 

I.  Commission  is  percentage,  (445). 

II.  The  sum  received  by  the  agent  as  the  price  of  property  sold, 
or  the  sum  invested  by  the  agent  in  the  purchase  or  exchange  of 
property,  is  the  base  of  commission,  (446). 


COMMISSION.  269 

III.  The  sum  remitted  to  an  agent,  and  including  both  the  pur- 
chase money  and  the  agent's  commission,  is  the  amount,  (4:417'). 

IV.  The  sum  due  the  employer  or  consignor  as  the  net  proceeds 
of  a  sale  or  collection,  is  the  difference,  (448). 

EXAMPLES    FOR    PRACTICE. 

1.  My  agent  sells  goods  to  the  amount  of  $6250 ;  what  is  his 
commission  at  3  %  ? 

OPER \TION  ANALYSIS.    According  to 

$6250  x  .03  =  $187.50  Prob-  J»  (449)>  AVG  multi- 

ply  the  sum  obtained  for 

the  goods,  $6250,  which  is  the  base  of  the  commission,  (II),  by  the 
rate  of  the  commission,  .03,  and  obtain  the  commission  or  percent- 
age, $187.50. 

2.  A  flour  merchant  remits  to  his  agent  in  Chicago  $3796,  for 
the  purchase  of  grain,  after  deducting  the  commission  at  4  %  ; 
how  much  will  the  agent  expend  for  his  employer,  and  what  wil) 
be  his  commission  ? 

OPERATION.  ANALYSIS.      Ac- 

1.00  +  .04  =  1.04  cording     to     Prob. 

$3796  -^  1.04  =  $3650,  for  grain,  IV,  (452),  we   di- 

$3796  —  $3650  =  $146,  commission.          vide  the  remittance, 

$3796,    which      is 

amount,  (HI),  by  1  plus  the  rate  of  commission,  or  1.04,  and  obtain 
the  base  of  commission,  $3650,  which  is  the  sum  to  be  expended  in 
the  purchase.  Subtracting  this  from  the  remittance,  we  have  $146, 
the  commission. 

NOTK  1. — It  is  evident  that  the  whole  remittance,  $3796,  should  not  be  taken 
as  the  base  of  commission  ;  for  that  would  be  computing  commission  on  commis- 
sion. A  person  must  charge  commission  only  on  what  he  expends  or  collects,  in 
his  capacity  as  agent. 

3.  A  factor  sold  real  estate  on  commission  of  5  <f0j  and  returned 
to  the  owner,  as  the  net  proceeds,  $8075 ;  for  what  price  did  he 
sell  the  property,  and  what  was  his  commission  ? 

OPERATION.  ANALYSIS.    According  to 

1.00 .05  =  .95  Prob-  Y>   (453),  we  divide 

$8075  -4-  .95  =  $8500,  price.          the    net    proceeds,   $8075, 
—  $8075  =  $425,  com.          which  is  difference,    (IV), 

by   1   minus    the   rate   of 
23* 


270  PERCENTAGE. 

commission,  and  obtain  the  base,  $8500,  which  is  the  price  of  the 
property  sold  ;  whence  by  subtraction,  we  obtain  the  commission, 
$425. 

4.  An  agent  sold  my  house  and  lot  for  $8600 ;  what  was  his 
commission  at  21  fy  ?  Ans.  $193.50. 

5.  A  lawyer    collects    $750.75 ;    what    is    his    commission    at 
3f  %?  Ans.  $28.15  +  . 

6.  My  agent  in  New  York  has  sold  3500  bushels  of  Indiana 
wheat  @  $1.40,  and  3600  bushels  of  dent  corn  @  $.74;  what  is 
his  commission  at  2^  %  ? 

7.  A  dealer  in  Philadelphia  sells  hides  on  commission  of  8J  %, 
as  follows:  2000  Ib.  Orinoco  @  $.23  J,  5650  Ib.  Central  Ameri- 
can @  $.22,  450  Ib.  Texas  @  $.23,  and  650  Ib.  city  slaughter 
@  $.21;  what  does  he  receive  for  ML  services  ?     Ans.  $162.75. 

8.  A  commission  merchant  sold  a  consignment  of  flour  and  pork 
for  $25372.     He  charged  $132  for  storage,  and  6i  %  commis- 
sion ;  what  were  the  net  proceeds  of  the  sale  ? 

9.  An  agent  for  a  Rochester  nurseryman  sells  4000  apple  trees 
at  $25  per  hundred,  2000  pear  trees  at  $50  per  hundred,  1600 
peach  trees  at  $20  per  hundred,  1800  cherry  trees  at  $50  per 
hundred,  and  500  plum  trees  at  $50  per  hundred  ;  what  is  his 
commission  at  30  %,  and  how  much  should  he  return  to  his  em- 
ployer as  the  net  proceeds,  after  deducting  $203.50  for  expenses? 

Ans.  Commission,  $1041  ;  Net  proceeds,  $2225.50. 

10.  A  lawyer  having  a  debt  of  $785  to  collect,  compromises 
for  82  %  ',  what  is  his  commission,  at  5  %  ?         Ans.  $32.185. 

11.  I  purchased  in  Chicago  4000  bushels  of  wheat  @  $1.25, 
and  shipped  the  same  to  my  agent  in  Oswego,  N.  Y.,  who  sold 
it  @  $1.50;  how  much  did  I  make,  after  paying  expenses  amount- 
ing to  $415,  and  a  commission  of  3  %  ?  Ans.  $405. 

12.  An  agent   received    $63  for  collecting  a  debt  of  $1260; 
what  was  the  rate  of  his  commission?  Ans.  5  %. 

13.  My  Charleston  agent  has  charged   $74.25  for  purchasing 
26400  Ib.  of  rice  at  $4.50  per  100  Ib. ;  required  the  rate  of  his 
commission. 

14.  A  house  and  lot  were  sold  for  $7850,  and  the  owner  re- 


COMMISSION.  271 

ceived  $7732.25  as  the  net  proceeds;  what  was  the  rate  of  com- 
mission ? 

15.  A  commission  merchant  in  Boston  having  received  28000 
Ib.  of  Mobile  cotton,  effects  a  sale  at  $.12£  per  pound.     After 
deducting  $35.36  for  freight  and  cartage,  $10.50  for  storage,  and 
his   commission,  he  remits  to  his  employer  $3252.89  as  the  net 
proceeds  of  the  sale ;  at  what  rate  did  he  charge  commission  ? 

Ans.  5 1  <f0. 

16.  The  net  proceeds  of  a  sale  were  $5635,  the  commission  was 
$115 ;  what  was  the  rate  of  commission  ? 

17.  An  agent  received  $22.40  for  selling  grain  at  a  commission 
of  4  t/o  ;  what  was  the  value  of  the  grain  sold  ?  Ans.  560. 

18.  My  attorney,  in  collecting  a  note  forme  at  a  commission  of 
8  %,  received  as  his  fee  $6.80  ;  what  was  the  face  of  the  note  ? 

19.  Sent  to  my  agent  in  Boston  $255,  to  be  invested  in  French 
prints  at  $.15  per  yard,  after  deducting  his  commission  of  2  %  ; 
how  many  yards  shall  I  receive ?  Ann.    1666:;. 

20.  John  Kennedy,  commission  merchant,  sells  for  Ladd  &  Co. 
860  barrels  of  flour  @  $7.50,  on  a  commission  of  2J   %.     He 
invests  the  proceeds  in  dry  goods,  after  deducting  his  commission 
of  1£   %   for  purchasing ;  how  many  dollars'  worth  of  goods  do 
Ladd  &  Co.  receive?  Ans.  $6105.81  +  . 

21.  A  commission  merchant,  whose  rate  both  for  selling  and 
investing  is  5  Cf0 ,  receives  24000  Ibs.  of  pork,  worth  6  cents,  and 
$3000  in  cash,  with  instructions  to  invest  in  a  shipment  of  cotton 
to  London.     What  will  be  his  entire  commission  ?     Ans.  $280. 

22.  A  speculator  received  $3290  as  the  net  proceeds  of  a  sale, 
after  allowing  a  commission  of  6  %  ;  what  was  the  value  of  the 
property?  Ans.  $3500. 

23.  The  net  proceeds  of  a  shipment  of  500  tons  of  pressed  hay, 
after  deducting  a  commission  of  3  <fc ,  and  $500  for  other  charges, 
were  $6290 ;  what  was  the  selling  price  per  ton  ? 

24.  I  send  a  quantity  of  dry  goods  into  the  country  to  be  sold 
at  auction,  on  commission  of  9  %.     What  amount  of  goods  must 
be  sold,  that  my  agent  may  buy  produce  with  the  avails,  to  the 
Value  of  $3500,  after  retaining  his  purchase  commission  of  4  %  ? 


272  PERCENTAGE 

NOTE  2.  —  $3500  plus  the  agent's  commission  equals  the  net  proceeds  of  the 
sale. 

25.  Having  sold  a  consignment  of  cotton  on   3    ^  commission, 
I  am  instructed  to  invest  the  proceeds  in  city  lots,  after  deducting 
my  purchase  commission  of  2  ffc.    My  whole  commission  is  $265; 
what  is  the  price  of  the  city  lots  ?  Ans.  $5141 . 

26.  What  tax  must  be  assessed  to  raise  $50000,  the  collector's 
commission  being  f  %  ?  Ans.  $50377. 83 -f. 

STOCKS. 

463.  A  Company  is  an  association  of  individuals  for  the 
prosecution  of  some  industrial  undertaking.     Companies  may  bo 
incorporated  or  unincorporated. 

464.  A  Corporation  is  a  body  formed  and  authorized  by  law 
to  act  as  a  single  person. 

465.  A  Charter  is  the  legal  act  of  incorporation,  and  defines 
the  powers  and  obligations  of  the  incorporated  body. 

466.  A  Firm  is  the  name  under  which  an  unincorporated 
company  transacts  business. 

NOTE. — A  private  banking  company,  or  a  manufacturing  or  commercial  firm 
is  also  called  a  House. 

467.  The  Capital  Stock  of  a  corporation  is  the  money  con- 
tributed and  employed  to  carry  on  the  business  of  the  company. 

468.  Joint  Stock  is  the  money  or  capital  of  any  company, 
incorporated  or  unincorporated. 

469.  Scrip  or  Certificates  of  Stock  are  the  papers  or  docu- 
ments issued  by  a  corporation,  giving  the  members  their  respective 
titles  or  claims  to  the  joint  capital. 

470.  A  Share  is  one  of  the  equal  parts  into  which  capital 
stock  is  divided.     The  value  of  a  share  in  the  original  contribu- 
tion of  capital  varies  in  different  companies;  in  bank,  insurance, 
and  railroad  companies  of  recent  organization,  it  is  usually  $100. 

471.  Stockholders  are  the  owners  of  stock,  either  by  original 
title  or  by  subsequent  purchase.     The  stockholders  constitute  the 
company. 

NOTRS.  — I.  The  capital  stock  of  any  corporation  is  limited  by  the  charter.  As 
»  general  rule,  only  a  portion  is  paid  at  the  time  of  subscription,  the  residue 
being  reserved  for  future  outlays  or  disbursements. 


STOCKS.  273 

2.  When  the  capital  stock  has  been  all  paid  in,  money  maybe  raised,  if  neces- 
sary, by  loans,  secured  by  mortgage  upon  the  property.     The  bonds  issued  for 
these  loans  entitle  the  holders  to  a  fixed  rate  of  interest, 

3.  Stocks,  as  a  general  name,  applies  to  the  scrip  and  bonds  of  a  corporation, 
to  government  bonds  and  public  securities,  and  to  all  paper  representing  joint 
capital  or  claims  upon  corporate  bodies. 

4.  The  members  of  an  incorporated  company  are  individually  liable  for  the 
debts  and  obligations  of  the  company,  to  the  amount  of  their  interest  or  stock 
in  the  company,  and  to  no  greater  amount.     But  the  members  of  a  firm  or  house 
are  individually  liable  for  all  the  debts  and  obligations  of  the  company,  "without 
regard  to  the  amount  of  their  share  or  interest  in  the  concern. 

The  calculations  of  percentage  in  stocks  are  treated  in  this  work 
under  the  heads  of 

Stock-jobbing,  Assessments  and  Dividends,  and  Stock  Invest- 
ments. 


STOCK-JOBBING. 

472.  Stock-jobbing-  is  the  buying  and  selling  of  stocks  with 
a  view  to  realize  gain  from  their  rise  and  fall  in  the  market. 

473.  The  Nominal  or  Par  value  of  stock  is  the  sum  for 
which  the  scrip  or  certificate  is  issued. 

474.  The  Market  or  Real  value  of  stock  is  the  sum  for 
which  it  will  sell. 

475.  Stock    is  At  Par  when   it   sells   for  its   first  cost,  or 
nominal  value. 

47G.  Stock  is  Above  Par,  at  a  premium  or  advance,  when  it 
sells  for  more  than  its  nominal  value. 

477.  Stock  is  Below  Par,  or  at  a  discount,  when  it  sells  for 
less  than  its  nominal  value. 

NOTE. — When  the  business  of  a  company  pays  large  profits  to  the  stock- 
holders, the  stock  will  be  worth  more  than  its  original  cost;  but  when  the  busi" 
ness  does  not  pay  expenses,  the  value  of  the  stock  will  be  less  than  its  original 
cost.  The  average  market  value  of  stock  generally  varies  directly  as  thb  rate 
of  profit  which  the  business  pays. 

478.  A  Stock  Broker  is  a  person  who  buys  *nd  sells  stocks, 
either  for  himself,  or  as  the  agent  of  another. 

NOTB. — A  person  employed  by  a  manufacturer,  wholesale  dealer,  or  commission 
merchant,  to  seek  customers  and  close  bargains,  at  or  from  his  place  of  business, 
is  called  a  broker,  of  the  class  or  kind  corresponding  to  his  business. 

479.  Brokerage  is  the  fee  or  compensation  of  a  broker. 

480.  The  calculations  in  stock-jobbing  are  based  upon  the 
following  relations : 


274  PERCENTAGE. 

I.  Premium,  discount,  and   brokerage  are  each  a  percentage, 
computed  upon  the  par  value  of  the  stock  as  the  base. 

II.  The  market  value  of  stock,  or  the  proceeds  of  a  sale,  is  the 
amount  or  difference,  according  as  the  sum  is  greater  or  less  than 
the  par  value. 

NOTE  1. — In  all  examples  relating  to  stocks,  $100  will  be  considered  a  share, 
unless  otherwise  stated. 

EXAMPLES    FOR   PRACTICE. 

1.  What  cost  54  shares  of  Heading  Railroad  stock,  at  4£  % 

premium  ? 

OPERATION.  ANALYSIS.      We    first 

$5400  X    .045  =  $243,  premium.        compute    the     premium 

$5400  -f  $243  =  5643,  Ans.  upon  the  par  value  of  the 

Or,  stock,  and  find  it  to  be 

5400  x  $1.045  =  $5643,   Ans.         $243  ;  adding  this  to  the 

$5400,  we  obtain  the  cost, 

or  market  value,  $5643.  Or,  since  every  dollar  of  the  stock  will  cost 
$1  plus  the  premium,  or  $1.045,  $5400  will  cost  5400  X  $1.045  = 
$5643. 

2.  What  do  I  receive  for  32  shares  of  telegraph  stock,  which  a 
broker  sells  for  me  at  15  %  discount  charging  £  %  brokerage? 

OPERATION.  ANALYSIS.       Adding 

.15  ~f     .0025  =     1525  the  rate  of  brokerage  to 

$1.00  —  $.1525  =  $.8475  proceeds  the  rate  of  discount'  we 

of  $1  of  stock.  have   -1525  5    hence  &1 

3200  X  $.8475  =  $2712,  Ans.  will  bring  $1— $.1525= 

$.8475,  and  $3200  will 
bring  3200  X  $.8475  =  $2712. 

3.  I  put  $35400  into  -the  hands  of  a  broker  to  be  invested  in 
Missouri  State  Bonds,  when  their  market  value  is  12  %  below  par; 
how  many  shares  shall  I  receive,  if  the  broker  charges  £  %  for 
his  services  ? 

OPERATION. 

$1.00  —  $.12    =  $.88,  market  value  of  $1. 
$  .88  +  $.00*  =  .885,  cost  of  $1. 
$35400  -f-  .885    =  $40000  ==  400  shares,  Ans. 
ANALYSIS.     Since  the  stock  is  12  %  below  par,  the  market  value  of 
$1  is  $.88  ;  adding  the  rate  of  brokerage,  we  find  that  every  dollar  of 


STOCKS.  275 

the  stock  will  cost  me  $.885.   Hence  for  $35400  the  broker  can  buy 
$35400  -r  .885  =  $40000  =  400  shares. 

NOTKS.  —  2.  The  rate  of  brokerage  in  New  York  city  has  been  fixed  by  cus- 
tom at  i  per  cent. 

3.  Since  brokerage  hns  the  same  base  as  the  premium  or  discount,  the  rate  of 
brokerage  may  always  be  combined  with  the  rate  of  premium  or  discount,  by 
addition  or  subtraction,  as  the  nature  of  the  question  may  require. 

4.  The  price  of  stock  is  usually  quoted  at  a  certain  per  cent,  of  the  face,  or 
nominal  value.    Thus  stock  at  4  'ft  above  par  is  quoted  at  104  <jo  }  stock  at  5  % 
below  par  is  quoted  at  95  %  ;  and  so  on. 

4.  What  is  the  market  value  of  15  Ohio  State  bonds  at  112  %  ? 

Ans.  $1680. 

5.  What  shall  I  realize  on  20  shares  of  Panama  railroad  stock 
at  135  %,  brokerage  at  If  %  ?  Ans.  82665. 

6.  My  agent  bought  for  me  120  shares  of  N.  Y.  Central  rail- 
road stock,  paying  80f  %,  and  charging  brokerage  at  ^  %  ;  what 
did  the  stock  cost  me  ?  Ans.  §9750. 

7.  What  cost  86  shares  in  the  Merchants'  Bank,  at  a  premium 
of  7£  %,  brokerage  J  %? 

8.  A  speculator  invested  $21910  in  shares  of  the  Harlem  rail- 
road, at  a  discount  of  601  %  '>  how  niany  shares  did  he  buy  ? 

9.  If  400  shares  of  the  Bank  of  Commerce  sell  for  $40150, 
what  is  the  rate  of  premium?  Ans.  f  %. 

10.  A  broker  receives  $48447  to  be  invested  in  bonds  of  the 
Michigan  Central  railroad,  at  94£  %  ;  how  much  stock  can  he 
buy,  allowing  1£  %  brokerage? 

11  My  agent  sells  830  barrels  of  Genesee  flour  at  $6  per  barrel, 
commission  5  %,  and  invests  the  proceeds  in  stock  of  the  Penn- 
sylvania Coal  Company,  at  82f  %,  charging  i  %  for  making  the 
purchase;  how  many  shares  do  I  receive  ?  Ans.  57. 

12.  I  purchased  18  shares  of  Ocean  Telegraph  stock,  par  value 
$500  per  share,  at  a  premium  of  2  % ,  and  sold  the  same  at  a  dis- 
count of  28  %  •  what  was  my  loss  ?  Ans.  $2700. 

NOTE  5.  —  The  rate  of  loss  is  .02.  +  .28  =  .30,  or  .30  % . 

13.  A  speculator  exchanged  $3600  of  railroad  bonds,  at  5  % 
discount,  for  27  shares  of  stock  of  the  Suffolk  Bank,  at  3   % 
premium,  receiving  the  difference  in  cash ;  how  much  money  did 
he  receive  ? 

14.  A  merchant  owning  525  shares  in  the  American  Exchange 


276  PERCENTAGE. 

Bank,  worth  104   %,  exchanges   them  for  United  States  bonds 
worth  105  %  ;  how  much  of  the  latter  stock  does  he  receive? 

15.  I  purchased  12  shares  of  stock  at  a  premium  of  5  %,  and 
sold  the  same  at  a  loss  of  $96 ;  what  was  the  selling  price  ? 

16.  Having  bought  $64000  stock  in  the  Cunard  Line,  at  2  % 
premium,  at  what  price  must  I  sell  it,  to  gain  $2560  ? 

.  Ans.  106  %. 

17.  A  speculator  bought  250  shares  in  a  Carson  Valley  mining 
company  at  103  %,  and  150  shares  of  the  Western  Railroad  stock 
at  95  °/o  ;  he  exchanged  the  whole  at  the  same  rates,  for  shares  in 
the  N.  Y.  Central  Railroad  at  80  %,  which  he  afterward  sold  at 
85  %.     How  much  did  he  gain?  Ans.  $2500. 

18.  I  purchased  stock  at  par,  and  sold  the  same  at  3  %  pre- 
mium, thereby  gaining  $750 ;  how  many  shares  did  I  purchase  ? 

19.  A  broker  bought  Illinois  State  bonds  at  103  %,  and  sold 
at  105  °/G.     His  profits  were  $240;  what  was  the  amount  of  his 
purchase?  An&.  $12000. 

20.  A  man  invested  in  mining  stock  when  it  was  4  %  above 
par,  and  afterward  sold  his  shares  at  5£  %  discount.     His  loss 
in  trade  was  $760;  how  many  shares  did  he  purchase? 

21.  I  invested  $6864  in  Government  bonds  at  106  %,  paying 
\\  %  brokerage,  and  afterward  sold  the  stock  at  112  %,  paying 
•1£  °/c  brokerage  ;  what  was  my  gain  ?  Ans.  $208. 

22.  How  much  money  must  be  invested  in  stocks  at  3  %  ad- 
vance, in  order  to  gain  $480  by  selling  at  7  %  advance  ? 

23.  How  many  shares  of  stock  must  be  sold  at  4  %  discount, 
brokerage  i  %,  to  realize  $4775?  Ans.  50. 


INSTALLMENTS,    ASSESSMENTS,    AND    DIVIDENDS. 

4  81.  An  Installment  is  a  portion  of  the  capital  stock  re- 
quired of  the  stockholders,  as  a  payment  on  their  subscription. 

48S.  An  Assessment  is  a  sum  required  of  stockholders,  to 
meet  the  losses  or  the  business  expenses  of  the  company. 

483.  A  Dividend  is  a  sum  paid  to  the  stockholders  from  the 
profits  of  the  business. 


STOCKS.  277 

484.  Gross  Earnings  are  all  the  moneys  received  from  the 
regular  business  of  the  company. 

485.  Net  Earnings  are  the  moneys  left  after  paying  expenses, 
losses,  and  the  interest  upon  the  bonds,  if  there  be  any. 

48O.  In  the  division  of  the  net  earnings,  or  the  apportion- 
ment of  dividends  and  assessments,  the  calculations  are  made  by 
finding  the  rate  per  cent,  which  the  sum  to  be  distributed  or  as- 
sessed bears  to  the  entire  capital  stock.  Hence, 

487'.  Dividends  and  assessments  are  a  percentage  computed 
upon  the  par  value  of  the  stock  as  the  base. 

EXAMPLES    FOR   PRACTICE. 

1.  The  Long  Island  Insurance  Company  declares  a  dividend 
of  6  %  ;  what  does  A  receive,  who  owns  14  shares  ? 

ANALYSIS.       According      to 

OPERATION.  449,    we    multiply    the    base, 

$1400  X  .06  =  $84  $1400,   by  the   rate,  .06,  and 

obtain  the  dividend,  $84. 

2.  A  canal  company  whose  subscribed  funds  amount  to  $84000, 
requires  an  installment  of  $6300 ;  what  per  cent,  must  the  stock- 
holders pay? 

OPERATION.  ANALYSIS.    According  to 

$6300  ~  84000  =  .07J  450,    we    divide    the    in- 

stallment, $6300,  which  is 
percentage,  by  the  base,  $84000,  and  obtain  the  rate,  .07£  =  7£  %. 

3.  A  man  owns  56  shares  of  railroad  stock,  and  the  company 
has  declared  a  dividend  of  8  %  ;  what  does  he  receive  ? 

Am.  $448. 

4.  I  own  $15000  in  a  mutual  insurance  company;  how  many 
shares  shall  I  possess  after  a  dividend  of  6  %  has  been  declared, 
payable  in  stock?  Ans.  159  shares. 

5.  The  Pittsburgh  Gas  Company  declares  a  dividend  of  15  %  ; 
what  will  be  received  on  65  shares  ? 

6.  A  received  $600  from  a  4   %   dividend ;  how  much  stock 
did  he  own  ?  Ans.  $15000. 


278  PERCENTAGE. 

7.  The  paid-in   capital  of  an  insurance  company  is  $536000. 
Its  receipts  for  one  year  are  $99280,  and  its  losses 'and  expenses 
are  $56400;  what  rate  of  dividend  can  it  declare  ?     Am.  8  %. 

8.  The  net  earnings  of  a  western  turnpike  are  $3616,  and  the 
amount  of  stock  is  $56000 ;  if  the  company  declare  a  dividend 
of  6  <fo,  what  surplus  revenue  will  it  have?  Ans.  $256. 

9.  The  capital  stock  of  the  Boston  and  Lowell  Railroad  Co. 
is  $1830000,  and  its  debt  is  $450000.     Its  gross  earnings  for  the 
year  1858  were  $407399,  and  its  expenses  $217621.     If  the  com- 
pany paid  expenses,  and  interest  on  its  debt  at  5|  %,  and  reserved 
$78,  what  dividend  would  a  stockholder  receive  who  owned  30 
shares?  Ans.  $270. 

10.  The  charter  of  a  new  railroad  company  limits  the  stock  to 
$800,000,  of  which  3  installments  of  10  %,  25  %,  and  35  %,  re- 
spectively, have  been  already  paid  in.    The  expenditures  in  the  con- 
etruction  of  the  road  have  reached  the  sum  of  $540,000,  and  the 
estimated  cost  of  completion  is  $400,000.     If  the  company  call  in 
the  final  installment  of  its  stock,  and  assess  the  stockholders  for  the 
remaining  outlay,  what  will  be  the  rate  %  ?  Ans.  17£. 

11.  The  Bank  of  New  York,  having  $156753.19  to  distribute 
to  the  stockholders,  declares  a  dividend  of  5J  %  ',  what  is  the 
amount  of  its  capital?  Ans.  $2,985,775  nearly. 

12.  The  passenger  earnings  of  a  western  railroad  in  one  year 
were  $574375.25,  the  freight  and  mail  earnings  were  $643672.36, 
the  whole  amount  of  disbursements   were   $651113.53,  and  the 
company  was  able  to  declare  a  dividend  of  8  %  ;  how  much  scrip 
had  the  company  issued?  Ans.  $7086676. 

13.  Having  received  a  stock  dividend  of  5  %,  I  find  that  I 
own  504  shares ;  how"  many  shares  had  I  at  first  ?       Ans.  480. 

14.  I  received  a  6  %  dividend  on  Philadelphia  City  railroad 
stock,  and  invested  the  money  in  the  same  stock  at  75  <f0.     My 
stock  had  then  increased  to  $16200 ;  what  was  the  amount  of  my 
dividend?  Ans.  $900. 

15.  A  ferry  company,  whose  stock  is  $28000,  pays  5  %  divi- 
dends  semi-annually.     The    annual    expenses   of  the    ferry   a*o 
$2950 ;  what  are  the  gross  earnings  ?  Ans.  $5750. 


STOCKS.  279 

STOCK  INVESTMENTS.* 

488.  The  net  earnings  of  a  corporation  are  usually  divided 
among  the  stockholders,  in  semi-annual  dividends.     The  income  of 
capital  stock  is  therefore  fluctuating,  being  dependent  upon  the  con- 
dition of  business ;  while  the  income  arising  from  bonds,  whether  of 
government  or  corporations,  is  fixed,  being  a  certain  rate  per  cent., 
annually,  of  the  par  value,  or  face  of  the  bonds. 

489.  Federal  or  United  States  Securities  are  of  two  kinds: 
viz.,  Bonds  and  Notes. 

Bonds  are  of  two  kinds. 

First,  Those  which  are  payable  at  a  fixed  date,  and  are  known 
and  quoted  in  commercial  transactions  by  the  rate  of  interest  they 
bear,  thus, :  U.  S.  6's,  that  is,  United  States  Bonds  bearing  6  % 
interest. 

Second,  Those  which  are  payable  at  a  fixed  date,  but  which  may 
be  paid  at  an  earlier  specified  time,  as  the  Government  may  elect. 
These  are  known  and  quoted  in  commercial  transactions  by  a  combi- 
nation of  the  two  dates,  thus  :  U.  S.  5-20 's,  or  a  combination  of  the 
rate  of  interest  and  the  two  dates,  thus :  U.  S.  6's  5-20 ;  that  is, 
bonds  bearing  6  °/0  interest,  which  are  payable  in  twenty  years,  but 
may  be  paid  in  five  years,  if  the  Government  so  elect. 

When  it  is  necessary,  in  any  transaction,  to  distinguish  from  each 
other  different  issues  which  bear  the  same  rate  of  interest,  this  is 
done  by  adding  the  year  in  which  they  become  due,  thus  :  U.  S.  5's 
of  71;  U.  S.  5's  of  74;  U.  S.  6's  5-20  of '84 ;  U.  S.  6's  5-20 
of  '85. 

Notes  are  of  two  kinds. 

First,  Those  payable  on  demand,  without  interest,  known  as  United 
States  Legal-tender  Notes,  or,  in  common  language,  "Green  Backs." 

*  The  following  eight  pages  contain/o«?*  pages  of  new  matter,  on  U.S.  Securities, 
Bonds,  Treasury  Notes,  Gold  Investments,  &c.,  to  meet  a  necessity  which  did  not 
exist  at  the  time  this  book  was  written. 

The  pupil  will  find  the  Cases,  Rules,  and  Operations  of  the  previous  editions 
essentially  the  same  in  this,  with  additional  examples,  and  other  matter,  which  may 
be  used  or  omitted;  so  that  the  present  may  be  used  with  the  previous  editions 
with  little  or  no  inconvenience. 


280  PERCENTAGE. 

Second,  Notes  payable  at  a  specified  time,  with  interest,  known 
as  Treasury  Notes.  Of  these,  there  are  two  kinds,  —  Six-per-cent. 
Compound-interest  Notes,  and  Notes  bearing  7^  %  interest,  the 
latter  known  and  quoted  in  commercial  transactions  as  7.30's. 

The  nomenclature  here  explained  is  the  one  used  in  commercial 
transactions,  which  involve  similar  securities  of  States  or  corporations 

The  interest  on  all  bonds  is  payable  in  gold. 

The  interest  on  notes  is  payable  in  Legal-tender  Notes. 

When  Bonds  or  Stocks  are  sold,  a  revenue  stamp  must  be  used 
equal  in  value  to  one  cent  on  each  $100,  or  fraction  of  $100,  of  their 
currency  value.  If  sold  by  a  broker,  this  is  charged  to  the  person 
for  whom  they  are  sold. 

The  following  are  the  principal  United  States  Securities :  — 

BONDS. 

U.  S.  6's  of  1867. 

U.  S.  6's  of  1868. 

U.  S.  6's  of  1880. 

U.  S.  6's  of  1881. 

U.  S.  5's  of  1871.      • 

U.  S.  5's  of  1874. 

U.  S.  5-20's,  due  in  1882,  interest  6$. 

U.  S.  5-20's,  due  in  1884,  interest  6%. 

U.  S.  5-20's,  due  in  1885,  interest  Q%. 

U.  S.  10-40's,  due  in  1904,  interest  6$. 

U.  S.  5's  (New),  1881. 

TJ.  S.  4i's  "   1886. 

U.  S.  4's   "   1901. 

Pacific  Eailroad  6's  of  1895. 

Pacific  Railroad  6's  of  1896. 

NOTES. 

Compound-interest  Notes  of  1867. 
Compound-interest  Notes  of  1868. 
7.30  Notes  of  1867. 
7.30  Notes  of  1868. 


STOCKS.  281 

CASE   I. 

49O.    To  find  what  income  any  investment  will  pro- 
duce. 

1.    What  income  will  be  obtained  by  investing  $6840  in  stock 
bearing  6  % ,  and  purchased  at  95  %  ? 

OPERATION.  ANALYSIS.    Wedi- 

$6840  -  .95  =  $7200,  stock  purchased.      vide  *"    investment; 

$7200  X  -06  =  $432,  annual  income.          $6840'  ^  the  cost  of 

Si,  and  obtain  $7200, 

the  stock  which  the  investment  will  purchase,  (452)-  And  since  the 
stock  bears  6  %  interest,  we  have  $7200  X  -06  =  $432,  the  annual 
income  obtained  by  the  investment.  Hence, 

RULE.  — Find  how  much  stock  the  investment  will  purchase,  and 
then  compute  the  income  at  the  given  rate  upon  the  par  value. 

EXAMPLES    FOR    PRACTICE. 

1.  The  trustees  of  a  school  invested   $35374.80  in  the  U.  S. 
5  °/0  bonds  as  a  teachers'  fund,  purchasing  the  stock  at  102^ 

if  the  salary  of  the  Principal  be  $1000,  what  sum  will  be  left  to  pay 
assistants?  Ans.  $725.60. 

2.  A  young  man,  receiving  a  legacy  of  $48000,  invested  one  half 
in  5  %  stock  at  95£  %,  and  the  other  half  in  6  %  stock  at  112  %, 
paying  brokerage  at  |  %  ;  what  annual  income  did  he  secure  from 
his  legacy?  Ans.  $2530. 

3.  I  have  32300  to  invest,  and  can  buy  New  York  Central  6's 
at  85  %,  or  New  York  Central  7 's  at  95  °f0  ;  how  much  more  prof 
itable  will  the  latter  be  than  the  former,  per  year  ? 

4.  A  owns  a  farm  which  rents  for  $411  45  per  annum.     If  he 
sell  the  same  for  $8229,  and  invest  the  proceeds  in  IT.  S.  5-20's  of 
'84,  at  105   °f0,  paying  ^  %   brokerage,  will  his  yearly  income  be 
increased  or  diminished,  and  how  much  !     Ans.  Increased  $56.55. 

5.  A  sold  $8700  of  U.  S.  5-20's  of  '84  at  104  %,  paying  for 
necessary  revenue  stamps,  and  invested  the  proceeds  in  U.  S.  10-40's 
at  94  % ,  brokerage  £  %  both  for  selling  and  buying.     Did  he  gain 
or  lose  by  the  exchange,  and  how  much  annually? 

Ans.  $45.62—. 


282  PERCENTAGE. 

CASE  II. 

491.     To  find  what  sum  must  be  invested  to  obtain  a 
given  income. 

I.  What  sum  must  be  invested  in  Virginia  5  per  cent,  bonds, 
purchasable  at  80  %,  to  obtain  an  income  of  $600  ? 

OPERATION.  AXALYSIS.     Since 

$600  -f-  .05  =  $12000,  stock  required.         Si  of  the  stock  will 
$1200  X  -80  =  $9600,  cost  or  investment,     obtain  $.05  income, 

to  obtain  $GOO  will 

require  $GOO  4-. 05  =  $12000,  (Case  1).  Multiplying  the  par  value 
of  the  stock  by  the  market  price  of  $1,  we  have  $12000  X  -80  — 
$9600,  the  cost  of  the  required  stock,  or  the  sum  to  be  invested. 
Hence  the 

RULE.     I.    Divide  the  given  income  by  the  °/0  which  the  stock 
pays  ;  the  quotient  will  be  the  par  value  of  the  stock  required. 

II.  Multiply  the  par  value  of  the  stock  by  the  market  value  of 
one  dollar  of  the  stock  ;  the  product  will  be  the  required  investment. 

EXAMPLES    FOR   PRACTICE. 

1.  If  Missouri  State  6's  are  16  %  below  par,  what  sum  must  be 
invested  in  this  stock  to  obtain  an  income  of  $960  ? 

2.  What  sum  must  I  invest  in  U.  S.  5-20's  of  '82  at  96}  %, 
brokerage  |  %,  to  secure  an  annual  income  of  $1500. 

Am.  $24250. 

3.  How  much  must  I  invest  in  U.  S.  7-30's,  at  106  %,  that  ray 
annual  income  may  be  $1752  ?  Ans.  $25440. 

4.  If  1  sell  $15600  U.  S.  10-40's  at  97  %,  and  invest  a  suf- 
ficient amount  of  the  proceeds  in  U.  S.'  5-20's  of  '85  at  107  %  to 
yield  an  annual^  income  of  $540,  and  buy  a  house  with  the  re- 
mainder, how  much  will  the  house  cost  me  ?  Ans.  $5502. 

5.  Charles  C.  Thomson,  through  his,  broker,  invested  a  certain 
sum  of  money  in  U.  S.  6's  5-20  at  107£  °f0,  and  twice  as  mucli  in 
U.  S.  10-40's  at  98i  %,  brokerage  in  each  case  .V  %.     His  in- 
come from  both  investments  was  $1674.     How  much  did  he  invest 
in  each  kind  of  stock  ? 

Ans.  First  kind,  $10692.     Second  kind,  $21384. 


STOCKS.  283 

CASE  m. 

492.  To  find  what  per  cent,  the  income  is  of  the  in- 
vestment, when  stock  is  purchased  at  a  given  price. 

l\  What  per  cent,  of  my  investment  shall  I  secure  by  purchasing 
the  New  York  7  per  cents,  at  105  %  ? 

ANALYSIS.     Since  $1  of  the  stock 

OPERATION.  will  cost  Si. 05,  and  pay  $.07,  the  in- 

.07  -f-  1.05  =  G§  %.  come  is  Tfo  =  6f  %  of  the  invest- 

ment.    Hence  the 

RULE.  Divide  the  annual  rate  of  income  which  the  stock  bears 
by  the  price  of  the  stock  /  the  quotient  will  be  the  rate  upon  the  in- 
vestment. 

EXAMPLES    FOR    PRACTICE. 

1 .  What  per  cent,  of  his  money  will  a  man  obtain  by  investing 
in  6  per  cent,  stock  at  108  %  ?  Am.  5£  %. 

2.  What  is  the  rate  of  income  upon  money  invested  in  6  per  cent, 
bonds,  purchased  at  a  discount  of  16  %  ?  Ans.  7-f  %. 

3.  Panama  railroad  stock  is  at  a  premium  of  34^  %,  and  the 
charge  for  brokerage  is  li  %  ;  what  will  be  the  rate  of  income  on 
an  investment  in  these  funds  if  the  stock  pays  a  dividend  of  Si-  % 
annually?  Ans.  6£  %. 

4.  Which  is  the  better  investment,  to  buy  5's  at  70  %,  or  6's 
at  80  %  ? 

5.  Which  is  the  more  profitable,  to  buy  8's  at  120  % ,  or  5's  at 
75  %  ? 

6.  What  is  the  rate  of  income  upon  money  invested  in  U.  S. 
7-30's  at  106  %  ?  Ans.  6f-£  %. 

7.  Which   is   tilt)   better  investment,   U.    S.   5-20's  of  '84  at 
108^  %,  or  U.  S.  10-40's  at  98  %,  and  how  much  per  cent,  per 
annum?  Ans.  U.  S.  5-20's,  ^/Trr  %• 

8.  If  a  man  invest  $10000  in  U.  S.  10-40's  at  98  %,  and  ex- 
changes them  at  par  for  U.  S.  7-30's  at  102  %,  what  is  his  rate  of 
income  ? 

9.  What  per  cent  of  his  money  will  a  man  gain  by  investing  in 
Pacific  Railroad  6's  at  105       ? 


284  PERCENTAGE. 

CASE   IV. 

493.  To  find  the  price  at  which  stock  must  be  pur- 
chased to  obtain  a  given  rate  upon  the  investment. 

1.  At  what  price  must  6  per  cent,  stocks  be  purchased  in  order 
to  obtain  8  °f0  income  on  the  investment  ? 

OPERATION.  ANALYSIS.      Since  $.06,  the  in- 

$  06  —  08  —  $75  come  °f  ^1  °f  the  stock'  is  8  °f°  of 

the  sum  paid  for  it,  we  have,  (449), 

$.06  -^-  .08  =  $75,  the  purchase  price.     Hence, 

RULE.  Divide  the  annual  rate  of  income  which  the  stock  bears 
by  the  rate  required  on  the  investment ;  the  quotient  will  be  the 
price  of  the  stock. 

EXAMPLES    FOB    PRACTICE. 

1.  What  must  I  pay  for  Government  5  per  cents.,  that  my  in- 
vestment may  yield  8  °/0  ?  Ans.  G2£-  %. 

2.  At  what  rate  of  discount  must  the  Vermont  6  per  cent,  bonds 
be  purchased  that  the  person  investing  may  secure  6£  %  upon  his 
money?  Ans.  4  %. 

3.  What  rate  of  premium  does  7  per  cent,  stock  bear  in  the  mar- 
ket when  an  investment  pays  6  %  ? 

4.  A  speculator  invested  in  a  Life  Insurance  Company,  and  re- 
ceived a  dividend  of  6  %,  which  was  8^  °/0  on  his  investment;  at 
what  price  did  he  purchase?  Ans.  72  °]0. 

5.  What  must  I  pay  for  U.  S.  10-40's,  that  my  investment  may 
yield  6  %fi  Ans.  83^  %. 

G.  What  rate  of  premium  does  U.  S.  6's  5-20  bear  in  market 
when  an  investment  pays  5  %  ? 

7.  At  what  rate  of  discount  must  U   S.  7-30's  be  purchased, 
that  the  investment  shall  yield  10  °/0  ? 

8.  What  must  I  pay  for  government  6's  of  '81,  that  my  invest- 
ment may  yield  7  %  ? 


STOCKS.  285 

GOLD  INVESTMENTS. 

493  a.  Currency  is  a  term  used  in  commercial  language,  First, 
To  denote  the  aggregate  of  Specie  and  Bills  of  Exchange,  Bank 
Bills,  Treasury  Notes,  and  other  substitutes  for  money  employed  in 
buying,  selling,  and  carrying  on  exchange  of  commodities  between 
Various  nations.  Second,  To  denote  whatever  circulating  medium  is 
used  in  any  country  as  a  substitute  for  the  government  standard.  In 
this  latter  sense,  the  paper  circulating  medium,  when  below  par,  is 
called  Currency,  to  distinguish  it  from  gold  and  silver.  If,  from  any 
cause,  the  paper  medium  depreciates  in  value,  as  it  has  done  in  the 
United  States,  gold  becomes  an  object  of  investment,  the  same  as 
stocks.  In  commercial  language,  gold  is  represented  as  rising  and 
falling  ;  but  gold  being  the  standard  of  value,  it  cannot  vary.  The 
variation  is  in  the  medium  of  circulation  substituted  for  gold ;  hence, 
when  gold  is  said  to  be  at  a  premium,  the  currency,  or  circulating 
medium,  is  made  the  standard,  while  it  is  virtually  below  par. 

CASE    I. 

To  change  gold  into  currency. 

1.  How  much  currency  can  be  bought  for  $150  in  gold  when 
gold  is  at  170  %  ? 

OPERATION.  ANALYSIS.     Since  a  dollar  of  gold  is 

81.70  X  150  =  $255         ™rth  S1'.7°  ™ I™'7'  ^  £"  ^ 

as  many  times  $170  of  currency  bought 

as  there  are  dollars  of  gold.     Therefore,  SI. 70  X  150  =  $255  is  the 
amount  of  currency  which  can  be  purchased  for  $150  in  gold. 

PtULE.  Multiply  the  value  of  one  dollar  of  gold  in  currency  by 
the  number  of  dollars  of  gold. 

2.  What  is  the  value,  in  current  funds,  of  $250.47  gold,  when 
gold  is  at  142J  %  ?  Ans.  $357.546  — . 

3.  If  a  person  holds  $6000  U.  S.  10-40's,  what  would  be  his  an- 
nual income  in  current  funds  if  gold  is  at  157  °fc  ?      Ans.  $471. 

4.  A  merchant  purchased  a  bill  of  goods  for  which  he  was  to  pay 
$7000  in  currency,  or  $5500  in  gold,  at  -his  option.     "Will  he  gain 
or  lose  by  accepting  the  latter  proposition,  gold  being  at  138£  %, 
and  how  much  in  currency?  Ans.  Lose  $617.50. 


286  PERCENTAGE. 

5.  Bought  broadcloth  @  $3  in  gold,  and  sold  the  same  @  $4  in 
currency.     Did  I  gain  or  lose  by  the  transaction,  and  how  much  per 
cent,  in  currency,  gold  being  at  140  %  ?  Ans.  Lost  4?,f  %. 

6.  A  broker  invested  $3000  of  gold  in  U.  S.  G's,  which  were 
worth  102  %  in  currency.     What  was  his  annual  income  from  the 
investment,  gold  being  at  134  %  ?  and  what  the  rate  per  cent.  ?  N 

Ans.  to  first,  $236.47,  gold  =  7} -f  %  currency. 

7.  A  gentleman  invested  $10,000,  current  funds,  in  U.  S,  5-20's 
of  '85,  at  104  %.     "What  will  be  his  annual  income  in  currency 
when  gold  is  at  137  %  ?  Ans.  $790.38TV 

CASE    II. 

To  change  currency  into  gold. 

1.  How  much  gold  can  be  purchased  for  $75  current  funds,  gold 
being  at  150  %  ? 

ANALYSIS.     A  dollar  of  gold  cost  $1.50 

OPERAfiON.  in   currency,    therefore    there    can    be    as 

$75  _i_  $1.50  —  50  many  dollars  of  gold  purchased  for  $75  in 

currency  as  $1.50  is  contained  times  in  $75. 

RULE.     Divide  the  amount  in  currency  by  the  price  of  gold. 

2.  What  is  the  value  in  gold  of  a  dollar  in  currency  when  gold 
is  at  203  %  ?  Ans.  $.49*%. 

3.  Gold  being  the  standard,  what  is  the  rate  of  discount  upon  cur- 
rent funds  when  gold  is  at  134  %,  150  %,  175  %,  180  %,  215  %, 
and  398  %  ?     Ans.  to  first ,  25f?  %•     Ans.  to  last,  74}£*  %. 

4.  What  is  gold  quoted  at  when  a  dollar  in  currency  is  worth  20 
cents  in  gold,  CO  cts.,  50  cts.,  15  cts.,  25  cts.,  and  72  cts.  ? 

5.  How  many  yards  of  cotton,  at  25  cts.  in  gold,  can  be  purchased 
for  $250  current  funds,  when  gold  is  at  175  %  ? 

Ans.  571^  yds. 

0.  Sold  $7800  7.30  Treasury  Notes,  at  105  %,  and  invested  the 
proceeds  in  gold  at  145  %,  with  which  I  bought  U.  S.  10-40's  at 
GO  °/0  in  gold.  Will  my  yearly  income  be  increased  or  diminished 
by  the  transaction,  and  how  much  in  gold?  Ans.  Increased  $78. 

7.  Which  ia  the  better  investment,  a  bond  and  mortgage  at  7  c/o^ 
or  U.  S.  10-40's,  gold  being  134  •  and  what  per  cent,  in  gold? 


PROFIT  AND  LOSS.  287 

PROFIT   AND   LOSS. 

.   Profit  and  Loss  are  commercial  terms,  used  to  express 
the  gain  or  loss  in  business  transactions. 

49*5.    Gains  and  losses  are  usually  estimated  at  some  rate  per 
cent,  on  the  money  first  expended  or  invested.      Hence 

I    Profit  and  loss  are  reckoned  as  percentage  upon  the  prime  or 
first  cost  of  the  goods  as  the  base. 

II.   The  selling  price  of  the  goods  is  amount  or  difference,  ac- 
cording as  it  is  greater  or  less  than  the  prime  cost. 

EXAMPLES    FOR   PRACTICE. 

1.  A  merchant  bought  cloth  for  83.25  per  yard)  and  gained 
8  °/0  in  selling ;  what  was  the  selling  price  ? 

OPEKATION.  ANALYSIS.    Multi- 

$3.25  X  .08  =  8.26,  advance  in  price.  Ptying     the     prime 

$3.25  -f-  .26  =  83.51,  selling  price.  cost,  $3.25,  which  is 

Qr  the    base    of    gain, 

$3.25  x  1.08  =  83.51,  selling  price.  $'    b?  uthe    *«£• 

.Uo,    we  have   ft. 26, 

the  gain,  which  added  to  the  cost  gives  $3.51,  the  selling  price.  Or, 
since  the  rate  of  gain  is  8  %,  that  which  cost  $1  vdl^  bring  $1.08, 
and  the  selling  price  will  be  1.08  times  the  buying  price.  Hence 
$3.25  X  1.08  =  $3.51,  the  selling  price. 

2.  A  jobber  invested  $2560  in  dry  goods,  and  realized  $384 
net  profit ;  what  was  the  rate  per  cent,  of  his  gain  ? 

OPERATION  ANALYSIS.     According   to 

$384  -r-  $2500  =:  15  %  (Prob.n,  450),  we  divide  the 

gain,  $384,  which  is  percent- 
age, by  the  cost,  $2560,  which  is  the  base,  and  obtain  15  =  15  % ,  the 
rate  of  gain. 

3.  A  produce  dealer  sold  a  shipment  of  wheat  at  a  loss  of  5  ^, 
realizing  as  the  net  proceeds,  $8170 ;  what  was  the  cost  ? 

OPERATION.  ANALYSIS      According  to 

gl  .00 .05  =  .95  (Prob.  V,  453),  we  divide  the 

8170  -f-  .95  =  $8600,  An*.          net  proceeds,  $8170,  which 

is  difference,  (448),  by  1 
minus  the  rate  of  loss,  or  .95,  and  obtain  the  base,  or  prime  cost,  $8600, 


288  PERCENTAGE. 

4.  A  merchant  pays  $7650  for  a  stock  of  spring  goods;  if  he 
sell  at  an  advance  of  20  %  upon  the  purchase  pri<?e,  what  will  be 
his  profits,  after  deducting  $480  for  expenses?         Ans.  $1050. 

5.  Bought  320  yards  of  calico  @  15  cents,  and  sold  it  at  a 
reduction  of  2£  %  ;  what  was  the  entire  loss? 

6.  A  dealer  having  bought  30  barrels  of  apples  at  $3.50  per 
barrel,  and  shipped  them  at  an  expense  of  $5.38,  to  be  sold  on 
commission  of  5   %,  what  will  be  his  whole  loss  if  the  selling 
price  is  10  %  below  the  purchase  price  ?  Ans.  $20.60^. 

7.  Bought  corn  at  $.50  a  bushel;  at  what  price  must  it  be  sold 
to  gain  33J  per  cent.  ? 

8.  Bought  fish  at  $4.25  per  quintal,  and  sold  the  same  at  $4.93 ; 
what  was  my  gain  per  cent.  ?  Ans.  16  <f0. 

9.  Bought  a  hogshead  of  sugar  containing  9  cwt.  44  lb.,  for  $59 ; 
paid  $4.72  for  freight  and  cartage;  at  what  price  per  pound  must 
it  be  sold  to  gain  20  per  cent,  on  the  buying  price  ? 

10.  A  wine  merchant  bought  a  hogshead  of  wine  for  $157.50 ; 
a  part  having  leaked  out  he  sold  the  remainder  for  $3.32£  a  gallon, 
and  found  his  loss  to  be  5  per  cent,  on  the  cost ;  how  many  gallons 
leaked  out?  Ans.  18. 

11.  Sold  a,  farm  of  106  A.  150  P.  for  $96  an  acre,  and  gained 
]  8  per  cent,  on  the  cost ;  how  much  did  the  whole  farm  cost  ? 

Ans.  $8700. 

12.  A  lumberman  sold  36840  feet  of  lumber  at  $21.12  per  M, 
and  gained  28  per  cent. ;  how  much  would  he  have  gained  or  lost, 
had  he  sold  it  at  $17  per  M?  Ans.  $18.42,  gained. 

13.  A  speculator  bought  shares  in  a  mining  company  when  the 
stock  was  4  %  below  par,  and  sold   the  same  when   it  was  28  °f0 
below  par;  what  per  cent,  did  he  lose  on  his  investment? 

14.  A  machinist  sold  a  fire  engine  for  $7050,  and  lost  5  per 
cent,  on  its  cost;  for  how  much  ought  he  to  have  sold  it  to  gain 
12  J  percent.?  Ans.  $8437.50. 

15.  Sold  my  carriage  at  30  per  cent,  gain,  and  with  the  money 
bought  another,  which  I  sold  for  $182,  and  lost  12^  per  cent.; 
how  much  did  each  carriage  cost  me  ?         .         (  First,  $160  ; 

i  Second,  $208. 


PROFIT  AND  LOSS.  289 

16.  Gaflfney  Burke  &  Co.  bought  a  quantity  of  dry  goods  for 
$6840;  they  sold  \  of  them  at  15  per  cent,  profit,  £  at  18|   per 
cent.,  \  at  20  per  cent,  and  the  remainder  at  33£  per  cent,  profit; 
how  much  was  the  average  gain  per  cent.,  and  how  much  the  whole 
gain  ?  Ans.   21|     %  gain  ;  $1482,  entire  gain. 

17.  If  I  buy  a  piece  of  land,  and  it  increases  in  value  each 
year  at  the  rate  of  50  per  cent,  on  the  value  of  the  previous  year, 
for  4  years,  and  then  is  worth  $12000,  how  much  did  it  cost? 

18.  A  Western  merchant  bought  wheat  as  follows  :  600  bushels 
of  red  Southern  @  $1.80,  1200  bushels  of  white  Michigan  @ 
$1.62^,    and    200  bushels   of  Chicago   spring,    @    $1.25.     He 
shipped  the  whole  to  his  correspondent  in  Buffalo,  who  sold  the 
first  two  kinds  at  an  advance  of  20  %  in  the  price,  and  the  bal- 
ance at  $1.20  per  bushel,  and  deducting  from  the  gross  avails  his 
commission  at  5  %,  and  $254.60  for  expenses,  returned  to  the 
consignor  the  net  proceeds.    What  was  the  rate  of  the  merchant's 
gain?  Ans.  4£  %. 

19.  A  broker  buys  stock  when  it  is  20  %  below  par,  and  sells 
it  when  it  is  16  %  below  par ;  what  is  his  rate  of  gain  ? 

20.  A  man  has  5  per  cent,  stock  the  market  value  of  which  is 
78  %  ;  if  he  sells  it,  and  takes  in  exchange  6  %  stock  at  4  % 
premium,  what  per  cent,  of  his  annual  income  does  he  lose  ? 

21.  A  machinist  sold  24  grain-drills  for  $125  each.     On  one 
half  of  them  he  gained  25  per  cent.,  and  on  the  remainder  he 
lost  25  per  cent.;  did  he  gain  or  lose  on  the  whole,  and  how 
much?  Ans.  Lost  $200. 

22.  Bought  land  at  $30  an  acre ;  how  much  must  1  ask  an  acre, 
that  I  may  abate  25  per  cent,  from  my  asking  price,  and  still  make 
20  per  cent,  on  the  purchase  money  ?  Ans.  $48. 

23.  A  salesman  asked  an  advance  of  20  per  cent,  on  the  cosi 
of  some  goods,  but  was  obliged  to  sell  at  20  per  cent,  less  than 
his  asking  price;  did  he  gain  or  lose,  and  how  much  per  cent.? 

24.  A  Southern  merchant  ships  to  his  agent  in  Boston,  a  quan- 
tity of  sugar  consisting  of  200  bbl.  of  New  Orleans,  each  containing 
216  lb.,  purchased  at  5  cents  per  pound,  and   560  bbl.  of  West 
India,  each  containing  200  lb.,  purchased  at  5f  cents  per  pound. 

25  T 


290  PERCENTAGE. 

The  agent's  account  of  sales  shows  a  loss  of  1  %  on  the  New  Or- 
leans, and  ?  profit  of  g|  %  on  the  West  India  sugar;  does  the 
merchant  gain  or  lose  on  the  whole  consignment,  and  what  .per 
cent.?  Ans.  Gains  f  %. 

25.  A  grocer  sold  a  hogshead  of  molasses  for  $31.50,  which 
v;as  a  reduction  of  30  %  from  the  prime  cost;  what  was  the  pur- 
chase price  paid  per  gallon  ? 

26.  A  speculator  sold  stock  at  a  discount  of  7|  %,  and  made 
a  profit  of  5     %  ;  at  what  rate  of  discount  had  he  purchased  the 
otock?  Ans.  12  %. 

27.  A  dry-goods  merchant  sells  delaines  for  2i  cents  per  yard 
more  than  they  cost,  and  realizes  a  profit  of  8  %  ;  what  was  the 
cost  per  yard?  Ans.   $.31}. 

28.  If  I  make  a  profit  of  18|  %  hy  selling  hroadcloth  for  $.75 
per  yard  above  cost,  how  much  must  I  advance  on  this  price  to 
realize  a  profit  of  31}  °/0  ? 

29.  A  speculator  gained  30  %  on  J  of  his  investment,  and  lost 
5  %   on  the  remainder,  and  his  net  profits  were  $720.     What 
would  have  been  his  profits,  had  he  gained  30  %  on  |  and  lost 
5  %  on  the  remainder?  Aus.  $405. 

30.  A  man  wishing  to  sell  his  real  estate  asked  36  per  cent, 
more  than  it  cost  him,  but  he  finally  sold  it  for  16  per  cent,  less 
than  his  asking  price.     He  gained  by  the  transaction  $740.48. 
How  much  did  the  estate  cost  him,  what  was  his  asking  price,  and 
for  how  much  did  he  sell  it  ? 

Ans.  Cost,  $5200;  asking  price,  $7072;  sold  for  $5940.48. 

31.  Sold  |  of  a  barrel  of  beef  for  what  the  whole  barrel  cost; 
what  per  cent,  did  I  gain  on  the  part  sold  ? 

32.  bought  4  hogsheads  of  molasses,  each  containing  84  gal- 
lons, at  $37i  a  gallon,  and  paid  $7.50  for  freight  and  cartage. 
Allowing  o  per  cent,  for  leakage  and  waste,  4  per  cent,  of  the  sales 
for  bad  debt.1,  and  1  per  cent,  of  the  remainder  for  collecting,  fo? 
how  much  per  gallon  must  I  sell  it  to  make  a  net  gain  of  25  per 
cent,  on  the  whole  cost?  Ans.  $.55-f-. 


INSURANCE.  291 

INSURANCE. 

Insurance  is  security  guaranteed  "by  one  party  to  ano- 
ther, against  loss,  damage,  or  risk.  It  is  of  two  kinds;  insurance 
on  property,  and  insurance  on  life. 

497.  The  Insurer  or  Underwriter  is  the  party  taking  the 
risk. 

498.  The  Insured  or  Assured  is  the  party  protected. 

499.  The  Policy  is  the  written  contract  between  the  parties. 
«5OO.    Premium  is  the  sum  paid  for  insurance.    It  is  always  a 

certain  per  cent,  of  the  sum  insured,  varying  according  to  the 
degree  or  nature  of  risk  assumed,  and  payable  annually  or  at  stated 
intervals. 

NOTKS. — 1.  Insurance  business  is  generally  conducted  by  joint,  stock  compa- 
nies, though  sometimes  by  individual*. 

2.  A  MntiKil  Innnrniicc  company  is  one  in  which  each  person  insured  is  enti- 
tled to  n  share  in  the  profit-?  of  the  concern. 

3.  The  act  of  insuring  is  sometimes  called  taking  a  risk. 

FIRE   AND    MARINE   INSURANCE. 

«5O1.  Insurance  on  property  is  of  two  kinds;  Fire  Insurance 
and  Marine  Insurance. 

Fire  Insurance  is  security  against  loss  of  property  by  fire. 

Marine  Insurance  is  security  against  the  loss  of  vessel  or  cai-go 
by  the  casualties  of  navigation. 

«5O2.  The  Sum  Covered  by  insurance  is  the  difference  be- 
tween the  sum  insured  and  the  premium  paid. 

NOTKS. — 1.  As  security  against  fraud,  most  insurance  companies  take  risks  at 
not  more  than  two  thirds  of  the  full  value  of  the  property  insured. 

2.  When  insured  property  suffers  damage  less  than  the  amount  of  the  policy, 
the  insurers  are  required  to  pay  only  the  estimated  loss. 

«>O«£.  The  calculations  in  insurance  are  based  upon  the  fol- 
lowing relations  : 

I.  Premium  is  percentage  (445)- 

II.  The  sum  insured  is  the  base  of  premium. 

III.  The  sum  covered  by  insurance  is  difference. 

EXAMPLES    FOR    PRACTICE. 

1.  What  premium  must  be  paid  for  insuring  my  stock  of  goods 
to  the  amount  of  $5760  at  U  %  ? 


292  PERCENTAGE. 

OPERATION.  ANALYSIS.     According  to 

85760  X  .0125  =  $72,  Ans.          Prob.  I,  (449),  we  multiply 

$5700,  the  base  of  premium, 
by  .0125,  the  rate,  and  obtain  $72,  the  premium. 

2.  For  what  sum  must  a  granary  be  insured  at  2  <J0  in  order  to 
cover  the  loss  of  the  wheat,  valued  at  $1617  ? 

OPERATION.  ANALYSIS.     According  to 

1.00 .02  =  .98  Prob-  V,  (453),  we  divide  tho 

$1617  -T-  .98  =  $1650,  Ans.          sum  to'  be  covered,   $1617, 

which   is    difference,    by    1 

minus  the  rate  of  premium,  and  obtain  $1650,  the  base  of  premium, 
or  the  sum  to  be  insured. 

PROOF.     $1650  X  .02  =  $33,  premium  ;  $1650  —  $33  =  $1617,  the 
sum  covered. 

3.  What  must  be  paid  for  an  insurance  of  $5860  at  1J  %  ? 

4.  What  is  the  premium  of  $360  at  £  %  ?  Ans.  $4.30. 

5.  What  is  the  premium  for  an  insurance  of  $3500  on  my  house 
and  barn,  at  H  %  ?  Ans.  $43.75. 

6.  A  fishing  craft,  insured   for  $10000  at  2i  %,  was   totally 
wrecked  •  how  much  of  the  loss  was  covered  ?        Ans.  $9775. 

7.  A  hotel  valued  at  $10000  has  been   insured  for  $6000  at 
li  %,  $5.50  being  charged  for  the  policy  and  the  survey  of  the 
premises ;  if  it  should  be  destroyed  by  fire,  what  loss  would  the 
owner  suffer  ?  Aits.  $4080.50. 

8.  A  merchant  whose  stock  in  trade  is  worth  $12000,  gets  the 
goods  insured  for  |  of  their  value,  at  f  %  ;  if  in  a  conflagration 
he  saves  only  $2000  of  the  stock,  what  actual  loss  will  he  sustain  ? 

9.  If  I  take  a  risk  of  $36000  at  2£  %,  and  re-insure  £  of  it 
at  3  %,  what  is  my  balance  of  the  premium?  Ans.  $360. 

10.  I  pay  $12  for  an  insurance  of  $800 ;  what  is  the  rate  of 
premium?  Ans.  \\  °f0. 

11.  A  trader  got  a  shipment  of  500  barrels  of  flour  insured  for 
80  %  of  its  cost,  at  31   %,  paying  $107.25  premium;  at  what 
price  per  barrel  did  he  purchase  the  flour?  Ana.  $8.25. 

12.  The  Astor  Insurance  Company  took  a  risk  of  $16000,  for 
a  premium  of  $280  ;  what  was  the  rate  of  insurance  ? 

13.  A  whaling  merchant  gets  his  vessel  insured  for  $20000  in 


INSURANCE.  293 

the  Gallatin  Company,  at  f  %,  and  for  $30000  in  the  Howard 
Company,  at  £  °/o  ;  what  rate  of  premium  does  he  pay  on  the  whole 
insurance?  Aus.  |  %. 

14.  If  it  cost  $46.75  to  insure  a  store  for  |  of  its  value,  at  If 
^,  what  is  the  store  worth?  Ans.  $6800. 

15.  For  what  sum  must  I  get  my  library  insured  at  1|  %,  to 
cover  a  loss  of  $7910  ?  Ans.  $8000. 

16.  What  will  be  the  premium  for  insuring  at  2|  %,  to  cover 
$27320?  Ans.  $680. 

17.  A  shipment  of  pork  was  insured  at  4f  %,  to  cover  f  of  its 
value.     The   premium  paid   was   $122.50;    what  was   the   pork 
worth?  Ans.  $4480. 

18.  A  gentleman  obtained  an  insurance  on  his  house  for  f  of 
its  value,  at  1J  %  annually.     After  paying  5  instalments  of  pre- 
mium, the  house  was  destroyed  by  fire,  in  consequence  of  which 
he  suffered  a  loss  of  $2940 ;  what  was  the  value  of  the  house  ? 

Ans.  $9600. 

19.  A  man's  property  is  insured  at  2J  ff0  payable  annually;  in 
how  many  years  will  the  premium  equal  the  policy  ? 

20.  A  company  took  a  risk  at  2i  %,  and  re-insured  |  of  it  in 
another  company  at  2f  %.     The  premium  received  exceeded  the 
premium  paid  by  $72.     What  was  the  amount  of  the  risk  ? 

21.  The  Commercial   Insurance  Company  issued  a  policy  of 
insurance  on  an  East  India  merchantman  for  f  of  the  estimated 
value  of  ship  and  cargo,  at  4  4  %  ,  and  immediately  re-insured  £ 
of  the  risk  in  the  Manhattan  Company,  at  3  %.    During  the  out- 
ward voyage  the  ship  was  wrecked,  and  the  Manhattan  Company 
lost  $1350  more  than  the  Commercial  Company;  what  did  the 
owners  lose?  Ans.  $40590., 

LIFE    INSURANCE.* 

5O-4.  Life  Insurance  is  a  contract  by  which  a  company  agrees 
to  pay  a  certain  sum   of  money  on  the  death   of  an  individual,  to 
his  heirs,  or  to  himself  if  he  survive  a  certain  number  of  years,  in 
*  This  entire  article  was  prepared  by  a  Professional  Actuary. 
25* 


294  PEKCEKTAGE. 

consideration  of  an  annual  premium  to  be  paid  during  life  or  for  a 
limited  number  of  years. 

«5O*5*  The  following  arc  the  principal  kinds  6f  policies  issued 
by  ii  Life  Insurance  Company  : 

1st.  Life  policies,  payable  at  the  death  of  the  person  insured,  the 
annual  premium  to  continue  during  life,  called  continued  premium 
life  policies. 

2d.  Life  policies,  payable  at  the  death  of  the  person  insured,  the 
annual  premium  to  continue  10  years,  5  years,  or  only  1  year,  called 
ten,  five,  or  single  payment  life  policies. 

3J.  Endowment  policies,  payable  to  the  person  insured,  at  tho 
end  of  a  certain  number  of  years,  as  ten,  fifteen,  twenty,  twenty- 
five,  thirty  or  thirty-five,  or  to  his  heirs  if  he  dies  sooner.  Annual 
premium  to  continue  during  the  existence  of  the  policy. 

4th.  Endowment  policies,  payable  as  the  preceding,  but  the  pay- 
ments all  to  be  made  in  one,  five,  or  ten  years. 

«5OO.  The  rates  of  premium  for  Life  Insurance,  as  fixed  by  dif- 
ferent Companies,  are  based  on  the  probabilities  of  life,  determined 
by  a  table  of  mortality,  and  the  probable  rates  of  interest  which 
money  will  bear,  and  a  loading  or  margin  for  expenses. 

5Q7.  A  Table  of  mortality  shows  how  many  persons  out  of  a 
given  number  (as  10000),  insuring  at  any  age.  may  bo  expected  to 
die  the  first,  second,  and  third  year,  and  so  on  until  they  are  all  dead. 

5O8o  The  premium  consists  of  three  elements: 

1st.  Reserve,  or  that  portion  of  each  premium  which  must  bo 
kept  and  improved  by  interest  (usually  four  per  cent.},  to  pay  the 
policy  at  its  certain  maturity. 

2d.  An  estimated  amount  for  each  man's  share  of  the  annual 
losses  of  the  Company. 

•3d.,  Loading,  or  a  certain  per  cent,  of  the  premium  to  meet  cur> 
rent  expenses. 

t>O9.  Most  Companies  in  this  country  are  mutual,  r.nd  divMo 
the  profits  among  the  policy  holders.  The  profits  result  from  the 
Company  realizing  upon  the  reserved  fun.l  more  than  the  assumed 
rate  of  interest,  four  per  cent.,  and  from  the  losses  by  death  being 
less  than  was  assumed  in  making  the  premium,  and  from  the  load- 
ing or  margin  being  more  than  the  expenses. 


IKSUKA^CE. 


295 


Dividends  are  declared  at  the  end  of  the  first,  second,  third,  or 
fourth  year,  and  may  be  applied  to  reduce  the  annual  premium  or 
to  increase  the  policy. 

NOTES.— 1.  One-half  of  the  premium  is  often  paid  by  a  note,  and  the  dividends 
are  af.cnvard  applied  toward  canceling  the  notes. 

2.  The  following  table  rates  have  been  selected  for  the  different  kinds  of 
policies,  for  the  reason  that  they  are  based  on  tin  American  Table  of  Mortality. 

8.  Stock  Companies  make  no  dividends  to  policy  holders,  but  generally  charge 
a  rate  of  premium  20  to  SO  per  cent,  less  than  the  'Mutual  Companies, 


LIFE  TABLE. 


AX.VUAL   PREMIUM    OX    A    POLICY   OK    $1000, 
Policy  parable  at  Death  only. 

Ago  nt    :         Payments 
Usue.     j       during   life. 

Payment 
for  10  years  only. 

Payment 
for  5  years  only. 

Single 
Payment. 

Age  at 

issue. 

C3 

8I9.S9 

$42.56 

$73.87 

$320.53 

25 

CO 

i'o  -n 

43.07 

75.25 

832.53 

26 

27 

20  0" 

44.22 

70.69 

833.83 

27 

23 

2143 

4,5.19 

73.13 

845.81 

23 

CD 

2_V,7 

40.02 

79.74 

852.05 

29 

83 

2.'  7;) 

4C.97 

81.30 

859.05 

SO 

81 

20..-:5 

47.93 

83.05 

8C6.8  '. 

81 

;  2 

24.C5 

43.C2 

84.83 

873.89 

82 

fiJJ 

Ci.73 

5.).  10 

83.G2 

881.73 

33 

84 

23.C6 

51.22 

8S.52 

SSXsS 

84 

83 

26.88 

52  4;) 

00.40 

898.  8  1 

a8) 

30 

27.23 

53.63 

92.54 

407.11 

86 

87 

2x17 

51.91 

94.07 

410.21 

37 

88 

20.15 

5C..C4 

95.SD 

425.64 

38 

ftD 

30.19 

57.63 

§9.19 

4:35  42 

39 

4) 

81.80 

69.09 

101.59 

415.55 

40 

41 

82.47 

60.60 

KJ4.0S 

456.04 

41 

4> 

88.72 

62.19 

106.C6 

466.89 

42 

43 

85.08 

63.84 

109.84 

478.11 

43 

44 

86.46 

65.57 

112.13 

489.71 

44 

45 

8797 

67.37 

115.02 

501.69 

45 

43 

89.58 

6926 

118.02 

514.04 

46 

47 

41.30 

71.25 

121.1.5 

526.78 

47 

43 

48.13 

73.32 

124.83 

539.88 

48 

4J 

46.09 

75.49 

127.74 

353.83 

49 

60 

47.18 

77.77 

131.21 

567.13 

50 

51 

49.40 

80.14 

13480 

5S1.24 

51 

5.! 

51.73 

82.63 

133.51 

595.  G6  r*. 

52 

M 

54.31 

8.5.22 

142.34 

610.36 

53 

54 

57  ()2 

S7.f4 

143.30 

625.88 

54 

55 

59.91 

"    90.79 

150.88 

640.54 

55 

;>6 

63.00 

C3  73 

1.54.60 

655.09 

56 

57 

66/29 

93.91 

158.04 

671.64 

57 

K 

C9.83 

100.2  1 

163.43 

687.48 

58 

59 

7-1  C;) 

103.  OS 

163.07 

703.49 

59 

00 

77.G:i 

107.3-5                  172.^7 

719  65 

60 

01 

61.96 

1  !  1  .23 

1  77.83 

785.02 

61 

69 

86.58 

115.82 

1S-2.T) 

7.52  26 

62 

c--> 

OL54 

119.06 

188.26 

768.67 

68 

64 

96  86 

124.28                  103.77 

785.10               64 

C5 

102.55 

129.18 

199.4S 

601.52 

65 

296 


PEKCEKTAGE. 


EKDOWMEKT  TABLE. 


ANNUAL  PREMIUM  TO  SECURE  flOOO,  payable  at  the  end  of  35,  30,  25,  etc.,  Years,  or  at  Death  if  prioj. 
Prem  urns  to  continue  until  Policy  matures. 

In 

In 

In 

In 

In 

In 

AliB. 

35  years. 

30  years. 

25  years. 

£0  years. 

15  years. 

10  years. 

AGE. 

25 
26 

$26.33 

26.57 

$30.61 

80.80 

$37.17 
37.34 

$47.68 
47.82 

$66.02 
60.15 

$108.91 

104.03 

25 

26 

27 

26.8=3 

81.02 

87.52 

47.98 

66.29 

104.16 

27 

28 

27.11 

31.  '25 

37.72 

48.15 

66.44 

104.29 

28 

29 

27.42 

31.50 

37.92 

4833 

CO.  CO 

104.43 

29 

80 

27.76 

81.78 

33.16 

48.53 

•    66.77 

1C4.58      1      30 

31 

28.13 

32.09 

38.41 

48.74 

60.06 

1(4.75      1     31 

32 

28.54 

32.43 

38.  69 

48.97 

(57.16 

1C-)  92    i    ;;2 

33 

28.93 

32.79 

£8.98 

49.22 

67.36 

1(5.11 

88 

34 

29.46 

83.19 

39.81 

49.49 

67.CO 

105.31 

34 

35 

30.  (iO 

83.  G3 

89.63 

49.79 

67.  S5 

K  '5:  3 

35 

36 

30.58 

34.11 

40.07 

50.11 

C8.12 

1(5.75 

86 

37 

81.22 

34.64 

40.50 

£0.47 

C8.41 

li.  (i  (  0 

87 

33 

81.93 

35.23 

40.93 

50.86 

08.73 

K  r,  -.8 

88 

39 

32.70 

85.  SS 

41.52 

51.30 

69.09 

106.58 

89 

4u 

33.55 

3659 

42.10 

51  .78 

6949 

106.90 

40 

41 

37.33 

42.75 

62.81 

69.92 

107.26 

41 

42 

38.24 

43.47 

62.89 

70.40 

107  (55 

42 

43 

39.19 

44.26 

68JB4 

70.92 

K  8.08 

43 

44 

40.23 

45.12 

54.25 

71.50 

118.55 

44 

45 

41.3T 

46.08 

65.04 

72.14 

li  9.07 

45 

46 

47.15 

55.91 

72.86 

1€965 

4$ 

47 

48.82      ' 

56.S9 

78.66 

110.30 

47 

4S 

49.61 

57.96 

74.54 

111.  (1 

48 

49 

51.04 

59.15 

75.51 

111  61 

49 

50 

52.60 

60.45 

76.59 

112.C8           £0 

51 

61.90 

77.77 

118.G4           M 

52 

63.48 

79.07 

114.70      i     .'2 

53 

05.22 

80.51 

li.r»6      ;      f« 

54 

67.14 

82.09 

117.14           M 

55 

69.24 

8:-!.  82 

118.C4 

56 

85.73 

120.09 

,r6 

57 

87.84 

121.78 

67 

58 

90.15 

1-::i;4 

59 

92.70 

126.70 

59 

60 

95.50 

127.96 

.60 

61 

13o  45 

61 

62 

188.19 

62 

63 

186.20 

68 

G4 

]:'.!'•  52 

C4 

G5 

143.16 

66 

EXAMPLES    FOR    PRACTICE. 

1.  What  sum  must  a  man  aged  33  pay  annually    for  life  for 
life  policy  of  $7500? 

"What  sum  annually  for  ten  years  ? 
AVhat  sum  annually  for  five  years? 
What  sum  in  a  single  payment  ? 


INSURANCE.  297 

OPERATION.  ANALYSIS.    We  multiply 

$24.78  x  7,500  =  $185.85,  Ans.  the  rate  per  thousand,  found 

50.10  x  7,500=     375.75,  Ans.  opposite  age  33,  Life  Table, 

86.02  x  7,500  =     649.65,  Ans.  by    the    number   of    thou- 

381.73  x  7,500  =2862.985,  Ans.  sands,  expressing  the  hun- 
dreds, tens,  and  units,  decimally. 

2.  A  man  30  years  of  age  takes  a  life  policy  in  a  Mutual  Com- 
pany, for  $5000,  the  premiums  continuing  until  death.     The  divi- 
dend reduces  the  annual    premium  an  average   of  30%.     He  dies 
after  making  21  payments;  how  much  more  money  will  his  family 
receive  than  he  has  paid  to- the  Company  ?  Ans.  $3331.55. 

3.  What  annual  premium  will  a  man  aged  36  years  pay  to  secure 
an  endowment  policy  for  $5000,  payable  to  himself  in  20  years,  or 
to  his  heirs  if  death  occurs  prior  ?  Ans.  $250.55. 

4.  A  young  man  aged  27  takes  an  endowment  policy  for  $'4000, 
payable  to  himself  in  25  years.  If  the  dividend  increases  his  policy 
$2400,  how  much  more  will  he  receive  than  he  has  paid  the  Com- 
pany ?  .  Ans.  $2648.00. 

5.  A  clergyman  aged  45  takes  an  endowment  policy  for  $3000, 
payable  to  himself  in   15  years,  or  to  his  family  at  death,  and  dies 
after  making   13   payments.      How  much  money  would   he  have 
saved  had  he  taken  a  policy  for  the  same  amount  on  the  continued 
payment  life  plan  ?  Ans.  $1332.63. 

6.  A  merchant  aged  49  insures  for  $8000  on  the  single  payment 
life  pl.m,  and  dies  in  the  fourth  year  thereafter.     How  much   less 
would  his  insurance  have  cost  him  had  he  insured  on  the  5  pay- 
ment life  plan  ?  Ans.  $338.96. 

7.  A  lawyer  aged  31  years  insures  in  the  Mutual  Life  Insurance 
Company  ofN.  Y.  for  $10000,  payable  to  himself  in  20  years.     If 
the  dividends  increase  the  policy  $7000,  how  much  more  will  he 
receive  than  he  has  paid  the  Company,  reckoning  Q%  simple  inter- 
est on  his  payments?  Ans.  $1110.70. 

NOTE.— 1st  payment  is  at  interest  20  years,  2d,  19  years,  3d,  18  years,  &c.,  &c, 

8.  A  has  his  life  insured  at  the  age  of  25  ;  B  insures  at  the  ago 
of  35,  each  taking  a  life  policy,  premiums  payable  until  death; 


298  PERCENTAGE. 

what  will   be  tho  age  of  each,  when  the  amount  of  premium  paid 
shall  exceed  the  face  of  the  policy  ?    Ans.  A,  75  yi»s. ;  B,  72  yrs. 

0.  A  person  at  age  of  34  had  his  life  insured   for  $0000,  pay- 
ments made  in  10  years.   When  he  died  there  was  a  net  gain  to  his 
family  of  $4463.40  ;   how  many  payments  had  he  made?     Ans.  5. 

10.  A  gentleman  aged  40  insures  his  life  in  the  Conn.  Mutual 
Life  Ins.  Co.  for  $5000,  premiums  to  continue  until  death.     After 
the*  fourth  year  his  premium  is  reduced  one-half  by  the  dividend. 
What  will  be  the  total  amount  of  premiums  paid  in  thirty  years? 

Ans.  $2660.50. 

11.  A  clergyman  insured  for  $5000  in  1843,  in  the  Mutual  Life 
Insurance   Company   of  New   York,    at    an    annual    premium    of 
$175.50,  andjdied  in  1885.     The  amount  paid  to  his  heirs,  includ- 
ing dividend  additions,  was  $8637.34.      How  much  better  WHS  this 
than  a  compound  interest  investment  at  6  %  ?         Ans.  $565.22. 

NOTE. — The  amount  oi'$l  per  annum  for  22  years  at  G%  comp.  inter,  is  $45.995. 

TAXES. 

510.  A  Tax  is  a  sum  of  money  assessed  on  the  person  or  pro- 
perty of  an  individual,  for  public  purposes. 

511.  A  Poll  Tax  is  a  certain  sum  required  of  each  male  citi- 
zen liable  to  taxation,  without  regard  to  his  property.    Each  person 
so  taxed  is  called  a  poll. 

512.  A  Property  Tax  is  a  sum  required  of  each  person  own- 
ing property,  and  is  always  a  certain  per  cent,  of  the  estimated 
value  of  his  property. 

£»!«$.  An  Assessment  Roll  is  a  list  or  schedule  containing  the 
names  of  all  the  persons  liable  to  taxation  in  the  district  or  com- 
pany to  be  assessed,  and  the  valuation  of  each  person's  taxable 
property. 

514L  Assessors  are  the  persons  appointed  to  prepare  the  as- 
sessment roll,  and  apportion  the  taxes. 

1.  In  a  certain  town  a  tax  of  $4000  is  to  be  assessed.     There 
are  400  polls  to  be  assessed  $.50  each,  and  the  valuation  of  the 
taxable  property,  as  shown  by  the  assessment  roll,  is  $950000; 
what  will  be  the  property  tax  on  $1,  and  how  much  will  be  A's 
tax,  whose  property  is  valued  at  $3500,  aud  who  pays  for  3  poih  ? 


TAXES. 


299 


OPERATION. 

8    .50  X     400  =  $200,  amount  assessed  on  the  polls. 
$4000  —  $200  =  $3800,  amount  to  be  assessed  on  property. 
$3800  -r-  $950000  =  .004,  rate  of  taxation  ; 
$3500  x  -004  =  $14,       A's  property  tax; 
8    .50x3    =        1.50,  A's  poll  tax; 

$15.50,  amount  of  A's  tax.     Hence  the 

RULE.  I.  Find  the  amount  of  poll  tax,  if  any,  and  subtract  it 
from  the  whole  tax  to  be  assessed  j  the  remainder  will  be  the  prop- 
erty tax. 

II.  Divide  the  property  tax  by  the  whole  amount  of  taxable 
property  ;  the  quotient  will  be  the  rate  of  taxation. 

III.  Multiply  each  man1  s  taxable  property  by  the  rate  of  taxa- 
tion, and  to  the  product  add  his  poll  tax,  if  any  ;   the  result  will  be 
the  whole  amount  of  his  tax. 

NOTE. — When  a  tax  is  to  be  apportioned  nmong  a  large  number  of  individuals, 
the  operation  is  greatly  facilitated  by  first  finding  the  tax  on  $1,  $2,  $3,  etc.,  to 
$9;  then  on  $10,  $20,  $30,  etc.,  to  $90,  and  so  on,  and  arranging  the  results  as 
in  the  following 

TABLE. 


Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

Prop. 

Tax. 

$1 

$.00i 

$10 

f.*4 

$100 

S  .40 

:  $1000 

$4.00 

2 

.003 

20 

.08 

200 

.80 

I  2000 

8. 

3 

.012 

30 

12 

300 

1.20 

3000 

12. 

4 

.016 

40 

.16 

400 

1.60 

4000 

16. 

5 

.020 

50 

.20 

500 

2.00 

5000 

29. 

6 

.024 

60 

.21 

600 

2.40 

6000 

24. 

7 

.028 

70 

.28 

700 

2.80 

7000 

28. 

8 

.032 

80 

.32 

800 

3.20 

8000 

32. 

9 

.030 

90 

.36 

900 

3.60 

9000 

36. 

EXAMPLES    FOR   PRACTICE. 


1.  According  to  the  conditions  of  the  last  example,  what  would 
be  the  tax  of  a  person  whose  property  was  valued  at  2465,  and 
who  pays  for  2  polls  ? 


300  PERCENTAGE. 

OPERATION. 

From  the  table  we  find  that 

The  tax  on  $2000  is  $8.00 

"      "      "        400  "  1.60 

n      u     u         60  (t  24 

«        a       a  fj  a  Q2 

And    "      ."      "  2  polls  "      1.00 

Whole  tax  "  $10.86,  Ans. 

2.  What  would  A's  tax  be,  who  is  assessed  for  $8530,  and  8 
polls?  Ans.  $35.62. 

3.  How  much  will  C's  tax  be,  who  is  assessed  for  $987,  and  1 
poll?  Ans.  $4.448. 

4.  The  estimated  expenses  of  a  certain  town  for  one  year  are 
$6319,  and  the  balance  on  hand  in  the  public  treasury  is  $854. 
There  are  2156  polls  to  be  assessed  at  $.25  each,  and  taxable  pro- 
perty to  the  amount  of  $1864000.     Besides  the  town  tax,  there 
is  a  county  tax  of  1?  mills  on  a  dollar,  and  a  State  tax  of  £  of  a 
mill  on  a  dollar.     Required  the  whole  amount  of  A's  tax,  whose 
property  is  valued  at  $32560,  and  who  pays  for  3  polls. 

5.  What  does  a  non-resident  pay,  who  owns  property  in  the 
same  town  to  the  amount  of  $16840  ?  Ans.  $79.99. 

6.  What  sum  must  be  assessed  in  order  to  raise  a  net  amount 
of  $5561.50,  and  pay  the  commission  for  collecting  at  2  %. 

NOTE.  —  Since  the  base  of  the  collector's  commission  is  the  sum  collected, 
(446),  the  question  is  an  example  under  Problem  V  of  Percentage. 

7.  In  a  certain  district  a  school  house  is  to  be  built  at  an  ex- 
pense of  $9120,  to  be  defrayed  by  a  tax  upon  property  valued  at 
$1536000.     What  shall  be  the  rate  of  taxation  to  cover  both  the 
cost  of  the  school  house,  and  the  collector's  commission  at  5  %  ? 

8.  The    expenses    of  a    school    for  one  term  were  $1200  for 
salary  of  teachers,  $57.65  for  fuel,  and  $38.25  for  incidentals ;  the 
money  received  from  the  school  fund  was  $257.75,  and  the  remain- 
ing part  of  the  expense  was  paid  by  a  rate-bill.     If  the  aggregate 
attendance  was  9568  days,  what  was  A's  tax,  who  sent  4  pupils  46 
days  each?  Ans.  $19.96+. 

9.  The  expense   of  building  a  public   bridge  was   $1260.52, 


GENERAL  AVERAGE.  301 

which  was  defrayed  "by  a  tax  upon  the  property  of  the  town.  The 
rate  of  taxation  was  3}  mills  on  one  dollar,  and  the  collector's 
commission  was  3£  %  ;  what  was  the  valuation  of  the  property  ? 

An*.  $401920. 


GENERAL  AVERAGE. 

5\5.  General  Average  is  a  method  of  computing  the  loss  to 
be  sustained  by  the  proprietors  of  the  ship,  freight,  and  cargo, 
respectively,  when,  in  a  case  of  common  peril  at  sea,  any  portion 
of  the  property  has  been  sacrificed  or  damaged  for  the  common 
safety. 

51  <5.  The  Contributory  Interests  are  the  three  kinds  of  prop- 
erty which  are  taxed  to  cover  the  loss.  These  are, 

1st    The  vessel,  at  its  value  before  the  loss. 

2d.  The  freight,  less  £  as  an  allowance  for  seamen's  wages. 

3d.  The  cargo,  including  the  part  sacrificed,  at  its  market  value 
in  the  port  of  destination. 

NOTE.  — In  New  York  only  £  of  the  freight  is  made  contributory  to  the  loss. 

5\Y.    Jettson  is  the  portion  of  goods  thrown  overboard. 

518.    The  loss  which  is  subject  to  general  average  includes, 

1st.  Jettson,  or  property  thrown  overboard. 

2d.  Repairs  to  the  vessel,  less  £  on  account  of  the  superior 
worth  of  the  new  articles  furnished. 

3d.  Expense  of  detention  to  which  the  vessel  is  subject  in  port. 

1.  The  ship  Nelson,  valued  at  $52000,  and  having  on  board  a 
cargo  worth  818000,  on  which  the  freight  was  $3600,  threw  over- 
board a  portion  of  the  goods  valued  at  $5000,  to  escape  wreck  in 
a  storm;  she  then  put  into  port,  and  underwent  repairs  amounting 
to  $1200,  the  expenses  of  detention  being  $350.  What  portion 
of  the  loss  will  be  sustained  by  each  of  the  three  contributing 
interests?  What  will  be  .paid  or  received  by  the  owners  of  the 
ship  and  freight  ?  What  by  A,  who  owned  $8000  of  the  car^o, 
including  $3500  of  the  portion  sacrificed,  and  by  B,  who  owned 
$6000  of  the  cargo,  including  $1500  of  the  portion  sacrificed,  and 
by  C,  who  owned  $4000,  or  the  residue  of  the  cargo  ? 
26 


H02  PERCENTAGE. 

OPERATION. 
LOSSES.  CONTRIBUTORY ,  INTERESTS. 

Jettson, $5000          Vessel, $52000 

Repairs,  less  £, 800          Freight,  less  £, 2400 

Cost  of  detention, ....       350          Cargo, 18000 

Total, $6150  Total, $72400 

$6150  -h     $72400  =  .0849447+,  rate  per  cent,  of  loss. 
$52000  X  .0849447  =  $4417.13,      payable  by  vessel. 
2400  X  .0849447  =      203.87,  "         "   freight. 

18000  X  .0849447  =  J.529.00,^  "        "   cargo. 

$6150.00,     Total  contribution. 

$8000  X  .0849447  ==  $679.56,  payable  by  A. 
0000  x  .0849447  =  509.67,  "  "  B. 
4000  x  .0849447  =  339.78,  "  "  C. 

$4417.13  +  $203.87  =  $4621.00,  payable  by  owners  of  vessel  and  freight. 

800.00  4-   350.00=   1150.00,       "      to 

4621.00—1150.00=   3471.00,  balance  payable  by  ship  owners. 
3500.00—679.56=2820.44,       "        receivable" by  A. 
1500.00—  509.67=     990.33,       "  "         "  B. 

Hence  the  following 

RULE.  I.  Divide  the  sum  of  the  losses  by  the  sum  of  the  con- 
tributory interests  ;  the  quotient  will  be  the  rate  of  contribution. 

II.  Multiply  each  contributory  interest  by  the  rate;  the  products 
will  be  the  respective  contributions  to  the  loss. 

EXAMPLES    FOR   PRACTICE. 

1.  The  ship  Nevada,  in  distress  at  sea,  cut  away  her  mainmast, 
and  cast  overboard  \  of  her  cargo,  and  then  put  into  Havana  to 
refit;  the  repairs  cost  $1500,  and  the  necessary  expenses  of  deten- 
tion were  8420.     The  ship  was  owned  and  sent  to  sea  by  Georgo 
Law,  and  was  valued  at  $25000 ;  the  cargo  was  owned  by  Hayden 
&  Co.,  and  consisted  of  2800  barrels  of  flour,  valued  at  $9  per 
barrel,  upon  which  the  freight  was  $4200.     In  the  adjustment  of 
the  loss  by  general  average,  how  much  was   due   from  Law  to 
Hayden  &  Co.?  An*.  $2629.36. 

2.  A  coasting  vessel  valued  at  $28000,  having  been  disabled  in 
a  storm,  entered  port,  and  was  refitted  at  an  expense  of  $270  for 
repairs,  and  $120   for  board  of  seamen,  pilotage,  and   dockage, 


CUSTOM  HOUSE  BUSINESS.  303 

Of  the  cargo,  valued  at  35000,  $2400  belonged  to  A,  $1850  to  B, 
and  6750  to  C ;  and  the  amount  sacrificed  for  the  ship's  safety 
was  §1400  of  A's  property,  and  8170  of  B's;  the  gross  charges 
for  freight  were  $1500.  Required  the  balance,  payable  or  re- 
ceivable, by  each  of  the  parties,  the  loss  being  apportioned  by 
general  average. 

,       f  §1295  payable  by  ship  owners;  $1268  receivable  by  A; 
S'|   41.25      "       «  C;  68.25        «         «  B. 

CUSTOM  HOUSE  BUSINESS. 

519.  Duties,  or  Customs,  are  taxes  levied  on  imported  goods, 
for  the  support  of  government  and  the  protection  of  home  industry. 

520.  A  Custom  House  is  an  office  established  by  government 
for  the  transaction  of  business  relating  to  duties. 

It  is  lawful  to  introduce  merchandise  into  a  country  only  at 
points  where  custom  houses  are  established.  A  seaport  town 
having  a  custom  house,  is  called  a  port  of  entry.  To  carry  on 
foreign  commerce  secretly,  without  paying  the  duties  imposed  by 
law,  is  smuggling. 

XOTK. — Customs  or  duties  form  the  principal  source  of  revenue  to  the  General 
Government  of  the  United  States:  by  increasing  the  price  of  imported  goods 
they  operate  as  an  indirect  tax  upon  consumers,  instead  of  a  general  direct  tax. 

521.  Duties  are  of  two  kinds  —  Ad  Valorem  and  Specific. 
Ad  Valorem  Duty  is  a  sum  computed  on  the  cost  of  the  goods 

in  the  country  from  which  they  were  imported. 

Specific  Duty  is  a  sum  computed  on  the  weight  or  measure  of 
the  goods,  without  regard  to  their  cost. 

522.  An  Invoice  is  a  bill  of  goods  imported,  showing  the 
quantity  and  price  of  each  kind. 

523.  By  the  New  Tariff  Act,  approved  March  2,  1857,  all 
duties  taken  at  the  U.  S.  custom  houses  are  ad  valorem.     The 
principal  articles  of  import  arc  classified,  and  a  fixed  rate  is  im- 
posed upon  each  list  or  schedule,  certain  articles  being  excepted 
and  entered  free. 

In  collecting  customs  it  is  the  design  of  government  to  tax 
only  so  much  of  the  merchandise  as  will  be  available  to  the  im- 


304  PERCENTAGE. 

porter  in  the  market.  The  goods  are  weighed,  measured,  gauged, 
or  inspected,  in  order  to  ascertain  the  actual  quantity  received  in 
port;  and  an  allowance  is  made  in  every  case  of  waste,  loss,  or 
damage. 

•531.  Tare  is  an  allowance  for  the  weight  of  the  box  or  the 
covering  that  contains  the  goods.  It  is  ascertained,  if  necessary, 
by  actually  weighing  one  or  more  of  the  empty  boxes,  casks,  or 
coverings.  In  common  articles  of  importation,  it  is  sometimes 
computed  at  a  certain  per  cent,  previously  ascertained  by  frequent 
trials  by  weighing. 

525.  Leakage  is  an  allowance  on  liquors  imported  in  casks 
or  barrels,  and  is  ascertained  by  gauging  the  cask  or  barrel  in 
which  the  liquor  is  imported. 

«>2G.  Breakage  is  an  allowance  on  liquors  imported  in 
bottles. 

527.  Gross  Weight  or  Value  is  the  weight  or  value  of  the 
goods  before  any  allowance  has  been  made. 

52S.  Net  Weight  or  Value  is  the  weight  or  value  of  the 
goods  after  all  allowances  have  been  deducted. 

NOTES.  —  1.  Draft  is  an  allowance  for  the  waste  of  certain  articles,  and  is 
made  only  for  statistical  purpone»  ;  it  does  not  affect  the  amount  of  duty. 

2.  Long  ton  me;i.«ure  is  employed  in  the  custom  houses  of  the  United  States, 
in  estimating  goods  by  the  ton  or  hundred  weight. 

The  rates  of  this  allowance  are  as  follows: 

On  112  lb lib. 

Above  112  lb.  and  not  exceeding  224  lb.,  2  lb. 
224  lb.  "  "  "  336  lb.,  3  lb. 

"   336  lb.  "  "    "    1120  lb.,  4  lb. 

"  1120  lb.  "  "     «    201(5  lb.,  7  lb. 

«  2010  lb 9  lb. 

529.  In  all  calculations  where  ad  valorem  duties  are  consid- 
ered, 

I.  The  net  value  of  the  merchandise  is  the  worth  of  the  net 
weight  or  quantity  at  the  invoice  price,  allowance  being  made  in 
cases  of  damage. 

II.  The  duty  is  computed  at  a  certain  legal  per  cent,  on  the 
net  value  of  the  merchandise. 

NOTE.  —  In  the  following  examples  the  legal  rates  of  duty,  according  to  the 
New  Tariff  Act,  are  given. 


CUSTOM  HOUSE  BUSINESS.  305 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  duty,  at  24  %,  on  an  invoice  of  cassimere  goods 

which  cost  $750  ? 

ANALYSIS.     According  to  Prob.  I, 
(449),  we  multiply  the  invoice,  $750, 

$750  X  .24  =  $180          which  is  the  base  of  the  duty,  by  the 
given  rate,  and  obtain  the  duty,  $180. 

2.  The  gross  weight  of  3  hogsheads  of  sugar  is  1024  lb.,  1016 
lb.,  and  1020  lb.  respectively;  t'ae  invoice  price  of  the  sugar  7j 
cents,  and  the  allowance  for  tare  80  lb.  per  hogshead;  what  is  the 
duty,  at  24  %  ? 

OPERATION.  ANALYSIS.   We  first  find 

1024  the   gross    weight   of   the 

1016  three    hhd.    from   which 

we  subtract  the  tare,  and 

3060,  gross  weight.        obtain  2820  lb.,   the  net 
80  X  3  =     240,  tare.  weight.    We  next  find  the 

.  ,  value  of  the   net  weight. 

2820,  net  weight.  „. 

a  Q-M  at    /£   cents,    the    invoice 

! L.  price,  and    then   compute 

$211.50,  net  value.  the  duty  at  24  %  on  this 

.24  value,  and  obtain  $50.76, 


$50.7600,  duty.  the  duty  required. 

3.  Having   paid  the  duty  at  8    %    on  a  quantity  of  Malaga 
raisins,  I  find  that  the  whole  cost  in  store,  besides  freight,  is  $378  ; 
what  were  the  raisins  invoiced  at  ? 

AJ^ALYSIS.      According   to   Prob. 

IV,   (452),  we  divide   the   amount, 
$378  ~  1.08  =  $350        $3T8j  by  i  plus  the  rate>  L08>  and 

obtain  the  base,  or  invoice,  $350. 

4.  A  Boston  jeweler  orders   from  Lubec  a  quantity  of  watch 
movements,  amounting  to  $2780 ;  what  will  be  the  duty,  at  4  %  ? 

5.  What  will  be  the  duty  at  15  %  on  1200  lb.  of  tapioca,  in- 
voiced at  5^  cents  per  pound  ?  Am.  §9.90. 

6.  What  is  the  duty  at  15  %  on   54   boxes  of  candles,  each 
weighing  1  cwt.,  invoiced  at  8|  cents  per  pound,  allowing  tare  at 
3^  per  cent.  ? 

26* 


506  PERCENTAGE. 

7.  A  merchant  imported  50  casks  of  port  wine,  each  contain- 
ing originally  3G  gallons,  invoiced  at  $2.50  per  gallon.      He  paid 
freight  at  $1.30  per  cask,  and  duty  at  30  %,  11  %  leakage  being 
allowed  at  the  custom  house,  and  $8.50  for  cartage ;  what  did  the 
•wine  cost  him  in  store  ?  Ann.  $5903.25. 

8.  A  liquor  dealer  receives  an  invoice  of  120  dozen  bottles  of 
porter,  rated  at  $1.25  per  dozen;  if  2    %  of  the  bottles  are  found 
broken,  what  will  bo  the  duty  at  24  %  ?  Am.  $35.28. 

9.  The  duty  at  19  %  on  an  importation  of  Denmark  satin  was 
$319.40,  what  was  the  invoice  of  the  goods?        Ans.  $3200. 

10.  The  duty  on  GOO  drums  of  figs,  each  containing  14  lb., 
invoiced  at  5|  cents  per  pound,  was  $35.28;  required,  the  rate 
of  duty.  Am.  8  %. 

11.  A  merchant  in  New  York  imports  from  Havana  200  hhd. 
of  Y\r.  I.  molasses,  each  containing  G3  gallons,  invoiced  at  8.30 
per  gallon;    150  hhd.  of  B.  coffee  sugar,  each   containing  500 
pounds,  invoiced  at  $.05  per  pound ;  80  boxes  of  lemons,  invoiced 
at  $2.50  per  box;  and  75  boxes  of  sweet  oranges,  invoiced  afe 
$3.00  per  box.     What  was  the  whole  amount  of  duty,  estimated 
at  24  %  on  molasses  and  sugar,  and  at  8  %  on  lemons  and  oranges  ? 

Ans.  $1841.20. 

12.  A  merchant  imported  5G  casks  of  wine,  each  containing  36 
gallons  net,  the  duty  at  30  %  amounting  to  $907.20 ;  at  what 
price  per  gallon  was  the  wine  invoiced  ? 

13.  The  duty  on  an  invoice  of  French  lace  goods  at  24  (fc,  was 
$132,  an  allowance  of  12   %   having  been  made  at  the  custom 
house  for  damage  received  since  the  goods  were  shipped ;  what 
was  the  cost  or  invoice  of  the  goods.  Ans.  $625. 

14.  A   quantity    of  Yalencias,    invoiced    at    $1G54,    cost    me 
$1980.50  in  store,  after  paying  the  duties  and  $12.24  for  freight; 
what  was  the  rate  of  duty  ? 

15.  The  duty  on  an  importation  of  Bay  rum,  after  allowing 
2  %  for  breakage,  was  $823.20,  and  the  invoice  price  of  the  rum 
was  $.25  per  bottle;  how  many  dozen  bottles  did  the  importer 
receive,duty  at  24  %  ?  Ans.  1143^  doz. 


SIMPLE  INTEREST. 


SIMPLE  INTEREST. 


307 


«53O.    Interest  u  a  rar.i  paid  for  the  use  cf  money. 
£>ol.    Principal  is  the  sum  fl-r  the  ucc  cf  -which  interest  13  paid. 
532.    Late  per  cent,  per  annum  is  the  sum  per  cent,  paid  f  jr  the 
use  cf  any  principal  for  one  year. 

NOTE.  —  The  rate  per  cent,  is  commonly  expressed  decimally  as  hundredth^  (<112). 

*5O3.    Amount  is  the  sum  of  iho  principal  and  interest. 

•>«M.  Simple  Interest  is  the  sum  paid  for  the  use  cf  Iho  princi- 
pal cnly,  during  the  whole  time  of  the  loan  or  credit. 

035.  Legal  Interest  is  the  rate  per  cent,  established  ly  law. 
It  varies  in  different  States,  as  follows  :  — 


Legal 
Rate, 
serceiit. 

Special  by 
Agreement. 

Legal 

Kate, 
per  cent. 

Special  by 
Agreement. 

Alabama  
Arkansas  
Ciliforaia  
Connecticut  .  . 
Colorado  

8 
6 

10 

7 
10 

c 

7 
8 

G 
8 
7 
10 
6 
6 
6 
7 
6 
5 
6 
6 
6 
7 
7 

Any  rate. 
Any  rate. 

Any  rate 
18  per  ct. 

10  per  ct. 
10  per  cf. 
10  per  ct 
Any  rate. 
10  per  ct. 
10  per  ct. 
10  per  ct. 
12  per  ct. 
10  per  ct. 
8  per  ct 
Any  rate 

Any  rate. 
10  per  ct. 
12  rer  ct. 

Mississippi  

Any  rate. 
1  0  per  ct. 
Any  rate. 

8  per  ct. 
15  per  ct. 
Any  rate. 
10  per  ct, 
12  per  ct. 

Any  rate. 
Any  late. 
10  per  ct. 
12  per  ct. 
Any  rate. 

12  per  ct. 

10  per  ct. 
Any  rate. 

Missouri 

Montana  

New  Hampshire... 
New  JorsL'C3T 

Cana  'a 

New  York  
North  Carolina  
Nebraska 

Dakota  
Delaware 

Dist.  Columbia 
Fioiida  

0'iio           .              . 

Oregon  .           .... 

Idaho 

Pennsylvania 

Illinois 

Rhode  Island 

Indiana  

South  Carolina  
Tennessee  
Texas  

Kansas  
Kentucky  
Louisiana  
Maine  
Maryland  
Massachusetts  . 
Miclii°"an  .  ... 

Utah  

Vermont           .  .  . 

Virginia  

West  Virginia  
W  iscon^in 

Washington  Ter..  . 
England.  . 

Minnesota.  . 

«5SG.    Usury  is  illegal  interest,  or  a  greater  per  cent,  than  tho 
Ic^al  rate. 

17oTK.  —  Tho  t:i!;ing  of  usury  is  prohibited,  under  various  penalties,  in  different 
States. 


308  PERCENTAGE. 


.    In  the  operations  of  interest  there  are  five  parts  or  ele- 
ments, namely  : 

I.  Rate  per  cent,  per  annum  ;  which  is  the  fraction  or  decimal 
denoting  how  many  hundredths  of  a  number  or  sum  of  money  are 
to  be  taken  for  a  period  of  1  year. 

II.  Interest;  which  is  the  whole  sum  taken  for  the  whole  period 
of  time,  whatever  it  may  be. 

III.  Principal  ;  which  is  the  base  or  sum  on  which  interest  is 
computed. 

IV.  Amount;  which  is  the  sum  of  principal  and  interest;  and 
Y.  Time. 

TO    COMPUTE    INTEREST. 
CASE    I. 

588.  To  find  the  interest  on  any  sum,  at  any  rate 
per  cent,  per  annum,  for  years  and  months. 

ANALYSIS.  In  interest,  any  rate  per  cent,  is  confined  to  1  year. 
Therefore,  if  the  time  be  more  than  1  year,  the  per  cent,  will  be  greater 
than  the  rate  per  cent,  per  annum,  and  if  the  time  be  less  than  1 
year,  the  per  cent,  will  be  less  than  the  rate  per  cent,  per  annum. 
From  these  facts,  we  deduce  the  following  principles  : 

I.  If  the  rate  per  cent,  per  annum  be  multiplied  by  the  time, 
expressed  in  years  and  fractions  or  decimals  of  a  year,  the  product 
will  be  the  rate  for  the  required  time.     And 

II.  If  the  principal  be  multiplied  by  the  rate  for  the  required 
time,  the  product  will  be  the  required  interest.     Hence 

III.  Interest  is  always  the  product  of  three  factors,  namely, 
rate  per  cent,  per  annum,  time,  and  principal. 

In  computing  interest  the  three  factors  may  be  taken  in  any  order  ; 
thus,  if  the  principal  be  multiplied  by  the  rate  per  cent,  per  annum, 
the  product  will  be  the  interest  for  1  year;  and  if  ihe  interest  for  1 
year  be  multiplied  by  the  time  expressed  in  years,  the  result  will  be 
the  required  interest.  Hence  the  following 

HULE.  I.  Multiply  the  principal  l>y  the  rate  per  cent.,  and  the 
product  will  Lfi  the  interest  for  1  year. 

II.  Multiply  this  product  l>y  the  time  in  years  and  fractions  of 
a  year;  the  result  will  be  the  required  interest. 


SIMPLE  INTEREST.  309 

Or,  Multiply  together  the  rate  per  cent,  per  annum,  time,  and 
principal,  in  such  order  as  is  most  convenient ;  the  continued  pro- 
duct will  be  the  required  interest. 

CASE   II. 

039.  To  find  the  interest  on  any  sum,  for  any  time, 
at  any  rate  per  cent. 

The  analysis  of  our  rule  is  based  upon  the  following 

Obvious  Relations  between  Time  and  Interest. 
I    The  interest  on  any  sum  for  1  year  at  1  per  cent.,  is  .01  of 
that  sum,  and  is  equal  to  the  principal  with  the  separatrix  re- 
moved two  places  to  the  left. 

II.  A  month  being  7V  of  a  year,  T12  of  the  interest  on  any  sum 
for  1  year  is  the  interest  for  1  month. 

III.  The  interest  on  any  sum  for  3  days  is  ^  =  T^  =  ,1  of  the 
interest  for  1  month,  and  any  number  of  days  may  readily  be  re- 
duced to  tenths  of  a  month  by  dividing  by  3. 

IV.  The  interest  on  any  sum  for  1  month,  multiplied  by  any 
given  time  expressed  in  months  and  tenths  of  a  month,  will  pro- 
duce the  required  interest. 

These  principles  are  sufficient  to  establish  the  following 

RULE.  I.  To  find  the  interest  for  1  yr.  at  1  %  : — Remove  the 
separatrix  in  the  given  principal  two  places  to  the  left, 

II.  To  find  the  interest  for  1  mo.  at  1  %  : — Divide  the  interest 
for  1  year  by  12. 

Ill  To  find  the  interest  for  any  time  at  1  %  :  —  Multiply  the 
•interest  for  1  month  by  the  given  time  expressed  in  months  and 
tenths  of  a  month. 

IV.  To  find  the  interest  at  any  rate  %  :  —  Multiply  the  interest 
at  1  °fo  for  the  given  time  by  the  given  rate. 

CONTRACTIONS.  After  removing  the  separatrix  in  the  principal  two 
places  to  the  left,  the  result  may  be  regarded  either  as  the  interest 
on  the  given  principal  for  12  months  at  1  per  cent.,  or  for  1  month  at 
12  per  cent.  If  we  regard  it  as  for  1  month  at  12  per  cent.,  and  if 
the  given  rate  be  an  aliquot  part  of  12  per  cent.,  the  interest  on  the 


310  PERCENTAGE. 

given  principal  for  1  month  may  readily  be  found,  by  taking  such  an 
aliquot  part  of  the  interest  for  1  month  as  the  given  rate  is  part  of  12 
per  cent.  Thus, 

To  find  the  interest  for  1  month  at  G  per  cent.,  remove  the  separa- 
trix  two  places  to  the  left,  and  divide  by  2. 

To  find  it  at  3  per  cent.,  proceed  as  before,  and  divide  by  4 ;  at  4 
per  cent,  divide  by  3  ;  at  2  per  cent.,  divide  by  G,  etc. 

SIX    PER    CENT     METHOD.* 

340.  By  referring  to  535  it  will  be  seen  that  the  legal  rate 
of  interest  in  22  States  is  6  per  cent.  This  is  a  sufficient  reason 
for  introducing  the  following  brief  method  into  this  work : 

Ax  A  LYSIS.     At  G  f0  per  annum  tho  interest  on  $1 

For  12  months is  $.00. 

"      2  months  (^=4 of  12  mo.) "    .01. 

"      1  month,  or  30  days  (T»§  of  12  mo.)  "     .OOJ  =  $.005  (TV  of  $.OG)% 

"      G  days  (i  of  30  da.).. ." "     .001. 

"      1     "     (i  of  G  da.  =  5\T  of  30  da.)  "     .000£. 
Hence  wo  conclude  that, 

1st.  The  interest  on  $1  is  $.005  per  month,  or  $.01  for  every 
2  months; 

2d.  Tho  interest  on  81  is  $.000£  per  day,  or  $.001  for  every  6 
days. 

From  these  principles  we  deduce  the 

RULE.  I.  To  find  the  rate: —  Call  every  year  $.06,  cv-ry  2 
months  $.01,  ever?/  G  days  $.001,  and  any  less  number  of  days 
sixths  of  1  mitt. 

II.  To  find  the  interest:  —  Multiply  the  principal  Ly  ike  rate. 

NOTES. — 1.  To  find  the  interest  nt  any  other  rate  <J0  by  this  method,  first  find 
it  at  6  ^,  and  tuen  increase  or  diminish  the  result  by  as  many  sixtlm.f  itself  na 
the  given  rate  is  units  greater  or  less  than  C  ^.  Thus,  lor  7  fi  add  £,  »or  -i  <J0 
subtract  \.  etc. 

2.  The  interest  of  $10  for  G  days,  or  of  $1  for  60  days,  is  $.01.    Therefore,  if  tho 
principal  be  less  than   $10  and   the  time  less  than  G  days,  or  the  principal  less 
than  SI  and  the  time  less  than  GO  days,  the  interest  will  be  less  than  $.01,  and 
may  be  disregarded. 

3.  Since  the  interest,  of  $1  for  GO  days  i?  $.01,  the  interest  of  $1  f>r  any  nunu 

*  This  method  of  finding  the  interest  on  $1  by  inspection  was  first  published 
in  The  Scholar's  Arithmetic,  by  Daniel  Adams,  M.  D.,  in  1801,  and  froiu  it» 
simplicity  it  has  come  into  very  general  use. 


SIMPLE  INTEREST.  311 

her  of  days  is  ns  many  cents  ns  CO  is  contained  times  in  the  number  of  day?. 
Therefore,  if  :iny  principal  be  multiplied  by  the  number  of  days  in  any  given 
number  of  months  and  days,  and  the  product  divided  by  CO.  the  result  will  lie 
the  interest,  in  cents.  That  is,  Multiply  the  j»rinrfjinf.  by  the  number  <>f  il"if*, 
divide  the  prmfnct  by  60,  find  point  off  ticn  ftecfmnl  place*  in  tlif  quotient.  Tlie 
retail  t  will  ue  the  interest  in  the  same  dfttotttKOtioH  as  the  principal. 

EXAMPLES    FOR   PRACTICE. 

What  is  the  interest  on  the  following  sums  for  the  times  given, 
at  6  per  cent.  1 

1.  $325  ror  3  years,  Ans.  858.50. 

2.  $1600  for  1  yr.  3  mo.  Ans.  6120. 

3.  $36.84  for  5  mo. 

4.  $35.14  for  2  yr.  9  mo.  15  da. 

5.  $217.15  for  3  yr.  10  mo.  1  da.  Ans.  $49.98+. 

6.  $721.53  for  4  yr.  1  mo.  18  da. 

7.  $15.125  for  15  mo   17  da.  Ans.  $1.17+. 
On  the  following  ac  7  per  cent.  ? 

8.  $2000  for  5  yr.  6  mo. 

9.  $1436.59  for  2  yr.  5  mo.  18  da.     Ans.  $248.051  +  . 

10.  $224.14  for  8  mo.  13  da.  Ans.  $11.026. 

11.  $100/25  for  63  da.  Ans.  $1.228+. 

12.  $600  for  24  da. 

13.  $520  for  5  yr.  11  mo.  29  da.  Ans.  $218.298. 

14.  $710.01  for  3  yr.  11  mo.  8  da. 
On  the  following  at  5  per  cent.  ? 

15.  $48  255  for  5  yr. 

16.  $750  for  1  yr.  3  mo. 

17.  $347.654  for  4  yr.  10  mo.  20  da.         Ans.  $158.315  +  . 

18.  $12850  for  90  da. 

19.  $2500  for  ^  mo.  20  da.  Ans.  $79.86- 

20.  $850.25  for  8  mo. 

21.  $48.25  for  1  yr.  2  mo.  17  da.  Ans.  $2.928  +  . 
On  the  following  at  8  per  cent.  ? 

22.  $2964.12  for  11  mo.  Ans.  $217.368  + 

23.  $725.50  for  150  da. 

24.  $360  for  2  yr  6  mo.  12  da. 

25.  §600  for  3  yr.  2  mo.  17  da.  Ana.  $154.266£. 


S12  PERCENTAGE. 

26.  $1700  for  28  da.  Ans.  $10.58. 
On  the  following  at  10  per  cent.  ? 

27.  83045.20  for  7  mo.  15  da.  Ans.1  $190.32  +  , 

28.  $1247.375  for  2  yr.  26  da.  Ans.  $258.48+.' 

29.  $2450  for  60  da. 

30.  $375.875  for  3  mo.  22  da. 

31.  $5000  for  10  da. 

32.  $127.65  for  1  yr.  11  mo.  3  da.  Ans.  $24.572. 

33.  What  is  the  interest  of  $155.49  for  3  mo.,  at  6i  per  cent.  ? 

34.  What  is  the  interest  of  $070.99  for  6  mo.,  at  5J  per  cent.  ? 

35.  What  is  the  amount  of  $350.50  for  2  yr.  10  mo.,  at  7  per 
cent.?  Ans.  $120.01+. 

36.  What  is  the  interest  of  $95.008  for  3  mo.  24  da.,  at  4-J  per 
cent.  ?  Ans.  $1.353. 

37.  What  is  the  amount  of  $145.20  for  1  yr.  9  mo.  27  da.,  at 
12  J  per  cent.  ?  Ans.  $178.32375. 

38.  What  is  the  amount  of  $215.34  for  4  yr.  6  mo.,  at  3  £  per 
cent.  ?  Ans.  $249.256. 

39.  What  is  the  amount  of  $5000  for  20  da.,  at  7  per  cent.  ? 

40.  What  is  the  amount  of  $16941.20  for  1  yr.  7  mo.  28  da., 
at  4|  per  cent.  ?  Ans.  $18277.91—. 

41    If  $1756.75  be  placed  at  interest  June  29,   1860,  what 
amount  will  be  due  Feb.  12,  1863,  at  7  %  1 

42.  If  a  loan  of  $3155.49  be  made  Aug.  15,  1858,  at  6  per 
cent.,  what  amount  will  be  due  May  1,  1866,  no  interest  having 
been  paid  ? 

43.  How  much  is  the  interest  on  a  note  for  $257.81,  dated 
March  1,  1859,  and  payable  July  16,  1861,  at  7  %  ? 

44.  A  person  borrows  $3754.45,  being  the  property  of  a  minor 
who  is  15  yr.  3  mo.  20  da.  old.     He  retains  it  until  the  owner  is 
21  years  old.     How  much  money  will  then  be  due  at  6  %  simple 
interest?  Ans.  $5037.22+. 

45.  If  a  person  borrow  $7500  m  Boston  and  lend  it  in  Wis- 
consin, how  much  does  he  gain  in  a  year? 

46.  A  man  sold  a  piece  of  property  for  $11320;  the  terms  were 
$3200  in  cash  on  delivery,  $3500  in  6  mo.,  $2500  in  10  mo.,  and 


SIMPLE  INTEREST.  313 

the  remainder  in  1  yr.  3  mo. ,  with  7  %  interest ;  what  was  the 
whole  amount  paid  ?  Ans.  $11773. 83£. 

47.  May  10,  1859,  I  borrowed  $6840,  with  which  I  purchased 
flour  at  $5.70  a  barrel.    June  21, 1860, 1  sold  the  flour  for  $6.62 J 
a  barrel,  cash.     How  much  did  I  gain  by  the  transaction,  interest 
being  reckoned  at  6  %  ? 

48.  If  a  man  borrow  $15000  in  New  York,  and  lend  it  in 
Ohio,  how  much  will  he  lose  in  146  days,  reckoning  360  days  toi 
the  year  in  the  former  transaction,  and  365  days  in  the  latter  ? 

49.  Hubbard  &  Northrop  bought  bills  of  dry  goods  of  Bowen, 
McNamee  &  Co.,  New  York,  as  follows,  viz.:   July  15,  1860, 
81250;  Oct.  4,  1860,  $3540.84;  Dec.  1,  1860,  $575;  and  Jan. 
24,  1861,  $816.90.     They  bought  on  time,  paying  legal  interest; 
how  much  was  the  whole  amount  of  their  indebtedness,  March  1, 
1861? 

50.  A  broker  allows  6  per  cent,  per  annum  on  all  moneys  de- 
posited with  him.     If  on  an  average  he  lend  out  every  $100  re- 
ceived on  deposit  11   times  during  the  year,  for  33   days  each 
time  at  2   %  a  month,  how  much  does  he  gain  by  interest  on 
$1000?  Ans.  $182. 

51.  A  man,  engaged  in  business  with  a  capital  of  $21840,  is 
making  12 }  per  cent,  per  annum  on  his  capital;  but  on  account  of 
ill  health  he  quits  his  business,  and  loans  his  money  at  7f  %. 
How  much  does  he  lose  in  2  yr.  5  mo.  10  da.  by  the  change? 

Ans.  $2535.861. 

52.  A  speculator  wishing  to  purchase  a  tract  of  land  containing 
450  acres  at  $27.50  an  acre,  borrows  th3  money  at  5J  per  cent. 
At  the  end  of  4  yr.  11  mo.  20  da.  he  sells  f  of  the  land  at  $34 
an  acre,  and  the  remainder  at  $32.55  an  acre.     How  much  does 
he  lose  by  the  transaction  ? 

53.  Bought  4500  bushels  of  wheat  at  $1.12  J  a  bushel,  payable 
in  6  months;  I  immediately  realized  for  it  $1.06  a  bushel,  cash, 
and  put  the  money  at  interest  at  10  per  cent.     At  the  end  of  the 
6  months  I  paid  for  the  wheat;  did  I  gain  or  lose  by  the  transac- 
tion, and  how  much  ? 

27 


814  PERCENTAGE. 

PARTIAL  PAYMENTS  OR  INDORSEMENTS. 

541.  A  Partial  Payment  is  payment  in  partf  of  a  note,  bond, 
or  other  obligation. 

542.  An  Indorsement  is  an  acknowledgment  written  on  the 
back  of  an  obligation,  stating  the  time  and  amount  of  a  partial 
payment  made  on  the  obligation. 

543.  To  secure  uniformity  in  the  method  of  computing  in- 
terest where  partial  payments  have  been  made,  the  Supreme  Court 
of  the  United  States  has  decided  that, 

I.  "  The  rule  for  casting  interest  when  partial  payments  have 
been  made,  is  to  apply  the  payment,  in  the  first  place,  to  the  dis- 
charge of  the  interest  then  due. 

II.  "  If  the  payment  exceeds  the  interest  the  surplus  goes  to- 
wards discharging  the  principal,  and  the  subsequent  interest  is  to 
be  computed  on  the  baiance  of  the  principal  remaining  due. 

III.  '  If  the  payment  be  less  than  the  interest  the  surplus  of 
interest  must  not  be  taken  to  augment  the  principal,  but  the  inte-. 
rest  continues  on  the  former  principal  until  the  period  when  the 
payments,  taken  together,  exceed  the  interest  due,  and  then  the 
surplus  is  to  be  applied  towards  discharging  the  principal,  and  the 
interest  is  to  be  computed  on  the  balance  as  aforesaid/' — Decision 
of  Chancellor  Kent. 

This  decision  has  been  adopted  by  nearly  all  the  States  of  the 
Union,  the  only  prominent  exceptions  being  Connecticut,  Ver- 
mont, and  New  Hampshire.  We  therefore  present  the  method 
prescribed  by  this  decision  as  the 

UNITED  STATES  RULE. 

I.  Find  the  amount  of  the  given  principal  to  the  tim:  of  the 
first  payment,  and  if  this  payment  exceed  the  interest  then  due, 
subtract  it  from  the  amount  obtained,  and  treat  the  remainder  as  a 
new  principal. 

II.  But  if  the  interest  be  greater  than  any  payment,  compute  the 
interest  on  the  same  principal  to  a  time  when  the  sum  of  the  jjcy- 
ments  shall  equal  or  exceed  the  interest  due,  and  subtract  the  sum 


PARTIAL  PAYMENTS.  315 

of  the  payments  from  the  amount  of  the  principal ;  the  remainder 
will  form  a  new  principal ,  with  which  proceed  as  before. 


EXAMPLES    FOR   PRACTICE. 


BUFFALO,  N   Y.,  May  15,  1856. 

1.  Two  years  after  date  I  promise  to  pay  to  David  Hudson,  or 
order,  one  thousand  dollars,  with  interest,  for  value  received. 

HENRY  BURR. 
On  this  note  were  indorsed  the  following  payments : 

Sept.  20,  1857,  received, $150.60 

Oct.    25,1859,       "        200.90 

July    11,  1861,       "        75.20 

Sept.  20,  1862,      "        112.10 

Dec.      5,1863,       "        105. 

What  remained  due  May  20,  1864? 

OPERATION. 

Principal  on  interest  from  May  15,  1856, $1000 

Interest  to  Sept.  20,  1857,  1  yr.  4  mo  5  da., 94.31 

Amount, $1094.31 

1st  Payment,  Sept  20,  1857, 150.60 

Remainder  for  a  new  principal, $943.71 

Interest  from  1st  paym't  to  Oct.  25, 1859,  2  yr.  1  mo.  5  da.,          138.54 

Amount, $1082.25 

2d  Payment,  Oct.  25,  1859, 200.90 

Remainder  for  a  new  principal, $881.35 

Interest  from  2d  paym't  to  Dec.  5, 1863,  4  yr.  1  mo.  10  da.,          253.63 

Amount, $1134.98 

3d  Payment,  less  than  interest  due, $75.20 

4th         "  112.11 

Sum  of  3d  and  4th  payments,  less  than  interest  due,  $187.31 
5th  payment, 105.00 

Sums  of  3d,  4th,  and  5th  payments, 292.31 

Remainder  for  new  principal $842.67 

Interest  to  May  20,  1864,  5  mo.  15  da., 27.04 

Balance  due  May  20,  1864, $869.71 


316  PERCENTAGE. 


$12QO-  RICHMOND,  VA.,  Oct.  15,  1859. 

2.  One  year  after  date  we  promise  to  pay  James  Peterson,  or 
order,  twelve  hundred  dollars,  for  value  received,  with  interest. 

WILDER  &  SON. 

Indorsed  as  follows:    Oct.  15,  1860,  $1000;   April  15,  1861, 
$200.    How  much  remained  due  Oct.  15, 1861  ?     Ans.  $82.56. 


$850TTjfe.  BOSTON,  June  10,  1855. 

3.  Eighteen  months  after  date  I  promise  to  pay  Crosby,  Nich- 
ols &  Co.,  or  order,  eight  hundred  fifty  and  y7^  dollars,  with 
interest,  for  value  received.  0.  L.  SANBORN. 

Indorsed  as  follows:  March  4,  1856,  $210.93;  July  9,  1857, 
$140;  Feb.  20,  1858,  $178;  May  5,  1859,  $154.30;  Jan.  17, 
1860,  $259  45.  How  much  was  due  Oct.  24,  1861  ? 


SAVANNAH,  GA.,  Sept.  4,  1860. 


4.  Six  months  after  date  I  promise  to  pay  John  Rogers,  or 
order,  three  hundred  eighty-four  and  T9^j  dollars,  for  value  re- 
ceived, with  interest.  WAI.  JENKINS. 

This  note  was  settled  Jan.  1,  1862,  one  payment  of  $126.50 
having  been  made  Oct.  20,  1861  ;  how  much  was  due  at  the  time 
of  •settlement  ? 

$3475.  NEW  ORLEANS,  March  6,  1857. 

5  On  demand  we  promise  to  pay  Evans  &  Hart,  or  order,  three 
thousand  four  hundred  seventy-five  dollars,  for  value  received,  with 
interest.  DAVIS  &  BROTHER. 

Indorsed  as  follows-  June  1,  1857,  $1247.60;  Sept  10,  1857, 
$1400.  How  much  was  due  Jan.  31,  1858  ? 

6.  A  gentleman  gave  a  mortgage  on  his  estate  for  $9750,  dated 
April  1,  1860,  to  be  paid  in  5  years,  with  annual  interest  after  9 
months  on  all  unpaid  balances,  at  10  per  cent.  Six  months  from 
date  he  paid  $846.50;  Oct.  20,  1862,  $2500;  July  3,  1863,  $1500; 
Jan.  1,  1864,  $500;  how  much  was  due  at  the  expiration  of  the 
given  time  ? 


PAKTIAL   PAYMENTS.  317 

$500.  PHILADELPHIA,  Feb.  1,  1861. 

7.  For  value  received,  I  promise  to  pay  J.  B.  Lippincott  &  Co., 
or  order,  five  hundred  dollars  three  months  after  date,  with  interest. 

JAMES  MONROE. 

Indorsed  as  follows:  May  1,  1861,  $40;  Nov.  14,  1861,.  88; 
April  1,  1862,  $12 ;  May  1,  1862,  $30.  How  much  was  due 
Sept.  16,  1862?  Ans.  $455.57-+. 

511.    CONNECTICUT  RULE.* 

I.  Payments  made  one  year  or  more  from  the  time  the  interest 
commenced,  or  from  another  payment,  and  payments  less  than  the 
interest  due,  are  treated  according  to  the  United  States  rule. 

II.  Payments  exceeding  the  interest  due  and  made  within  one 
year  from  the  time  interest  commenced,  or  from  a  former  payment, 
shall  draw  interest  for  the  balance  of  the  year,  provided  the  interval 
does  not  extend  beyond,  the  settlement,  and  the  amount  must  be  sub- 
tracted from  the  amount  of  the  principal  for  one  year;  the  re- 
mainder will  be  the  new  principal. 

III.  If  the  year  extend  beyond  the  settlement,   then  find  the 
amount  of  the  payment  to  the  day  of  settlement,  and  subtract  it 
from  the  amount  of  the  principal  to  that  day  ;  the  remainder  will 
be  the  sum  due. 

*>  1»5.  A  note  containing  a  promise  to  pay  interest  annually 
is  not  considered  in  law  a  contract  for  any  thing  more  than  simple 
interest  on  the  principal.  For  partial  payments  on  such  notes 
the  following  is  the 

VERMONT  RULE. 

I.  Find  the  amount  of  the  principal  from  the  time  interest  com- 
menced to  the  time  of  settlement. 

II.  Find  the  amount  of  each  payment  from  the  time  it  was  made 
to  the  time  of  settlement. 

III.  Subtract  the  sum  of  the  amounts  of  the  payments  from  the 
amount  of  the  principal;    the  remainder  will  be  the  sum  due. 

*  The  United  States  rule  is  in  general  use. 

27* 


318  PERCENTAGE. 


In  New  Hampshire  interest  is  allowed  on  the  annual 
interest  if  not  paid  when  due,  in  the  nature  of  .damages  for  its 
detention  ;  and  if  payments  are  made  before  one  year's  interest 
has  accrued,  interest  must  be  allowed  on  such  payments  for  the 
balance  of  the  year.  Hence  the  following 

NEW  HAMPSHIRE  RULE. 

I.  Find  the  amount  of  the  principal  for  one  year,  and  deduct 
from  it  the  amount  of  each  payment  of  that  year,  from  the  time 
it  was  made  up  to  the  end  of  the  year  ;  the  remainder  will  be  a 
new  principal,  with  which  proceed  as  before. 

II.  If  the  settlement  occur  less  than  a  year  from  the  last  annual 
term  of  interest,  make  the  last  term~of  interest  a  part  of  a  year, 
accordingly. 

EXAMPLES    FOR   PRACTICEc 

$1QQQ-  NEW  HAVEN,  CONN.,  Feb.  1,  1856. 

1.  Two  years  after  date,  for  value  received,  I  promise  to  pay  to 
Peck  &  Bliss,  or  order,  one  thousand  dollars  with  interest. 

JOHN  CORNWALL. 

Indorsed  as  follows:  April  1,  1857,  $80;  Aug.  1,  1857,  $30; 
Oct.  1,  1858,  $10  ;  Dec.  1,  1858,  $600  ;  May  1,  1859,  $200.  How 
much  was  due  Oct.  1,  1859  ?  Ans.  $266.38. 


BURLINGTON,  VT.,  May  10,  1858. 

2.  For  value  received,  I  promise  to  pay  David  Camp,  or  order, 
two  thousand  dollars,  on  demand,  with  interest  annually. 

RICHARD  THOMAS. 

On  this  note  were  indorsed  the  following  payments  :  March  10, 
1859,  $800;  May  10,  1860,  $400;  Sept.  10,  1861,  $300.  How 
much  was  due  Jan.  10,  1863  ? 

3.  How  much  would  be  due  on  the  above  note,  computing  by 
the  Connecticut  rule  ?  Am.  $831.58. 

4.  How  much,  computing  by  the  New  Hampshire  rule  ?     By 
the  United  States  rule  ?  .        (  N.  H.  rule,  $833.21 ; 

'  I U.  S.     «     $831.90. 


SAVINGS   BANK    ACCOUNTS.  319 

SAVINGS   BANK    ACCOUNTS. 

.  Savings  Banks  are  institutions  intended  to  receive  in 
trust  or  on  deposit,  small  sums  of  money,  generally  the  surplus 
earnings  of  laborers,  and  to  return  the  same  with  a  moderate  interest 
at  a  future  time. 

5>48.  It  is  the  custom  of  all  savings  banks  to  add  to  each 
depositor's  account,  at  the  end  of  a  certain  fixed  term,  the  interest 
due  on  his  deposits  according  to  some  general  regulation  for  allow- 
ing interest.  The  interest  term  with  some  savings  banks  is  6 
months,  with  some  3  months,  and  with  some  1  month. 

£519.  A  savings  bank  furnishes  each  depositor  with  a  book, 
in  which  is  recorded  from  time  to  time  the  sums  deposited  and 
the  sums  drawn  out.  The  Dr.  side  of  such  an  account  shows  the 
deposits,  and  the  Cr.  side  the  depositor's  checks  or  drafts.  In  the 
settlement,  interest  is  never  allowed  on  any  sum  which  has  not 
been  on  deposit  for  a  full  interest  term.  Hence,  to  find  the 
jnount  due  on  any  depositor's  account,  we  have  the  following 

RULE.  At  the  end  of  each  term,  add  to  the  balance  of  the 
account  one  term's  interest  on  the  smallest  balance  on  deposit  at  any 
one  time  during  that  term  ;  the  final  balance  thus  obtained  will  be 
the  sum  due. 

NOTES.  —  1.  It  will  be  seen  that  by  this  rule  no  interest  is  allowed  for 
money  on  deposit  during  a  partial  term,  whether  the  period  be  the  first  or  the 
last  part  of  the  term. 

2.  An  exception  to  this  general  rule  occurs  in  the  practice  of  some  of  the 
savings  banks  of  New  York  city.  In  these,  the  interest  term  is  6  months,  and 
the  depositor  is  allowed  not  only  the  full  term's  interest  on  the  smallest  balance, 
but  a  half  terra's  interest  on  any  deposit,  or  portion  of  a  deposit  made  during 
the  first  3  months  of  the  term,  and  not  drawn  out  during  any  subsequent  part  of 
the  term. 

EXAMPLES   FOR   PRACTICE. 

1.  What  will  be  due  April  20,  1860,  on  the  following  account, 
interest  being  allowed  quarterly  at  6  per  cent,  per  annum,  the 
terms  commencing  Jan.  1,  April  1,  July  1,  and  Oct.  1  ? 

Dr.       Savings  Bank  in  account  with  James  Taylor.       Cr. 

1858,  Jan.    12, $75         1858,  March  5, $30 

"       May  10, 150  "       Aug.  16, 50 

"       Sept     1, 20  «       Dec.     1,... 48 

1859,  Feb.  16, 130 


120 


PERCENTAGE. 


OPERATION. 

Deposit,  Jan.  12,  1858,  ........................  ......  $75 

Draft,  March  5,       "     ..........................  .1..     30 

Balance,  Apr.  1,  I860,  ..................  $45 

Deposit,  May  10,  1858,  ..............................   150 

Int.  on  $4^,  for  3  mo  .............  .  ....................  68 

Balance,  July  1,  1860,  ..................  $195.68 

Draft,  Aug.  16,  1858,  ......  .  .........................     50 

Least  balance  during  the  current  term,  ...........  $145.68 

Deposit,  Sept.  1,  1858,  ..............................     20.00 

Int.  on  $145.68,  for  3  mo  ...........................       2.19 

Balance,  Oct.  1,  1858,  ...................  $167.87 

Draft,  Dec.  1,  1858,  ..................................     48 

Least  balance  during  the  current  term,  ...........  119.87 

Int.  on  $119.87,  for  3  mo  ................  ,  .........       1.80 

Balance,  Jan.  1,  1860,  ...................  $121.67 

Deposit,  Feb.  16,1860,  ..............................   130.00 

Int.  on  $121.67,  for  3  mo  ...........................       1.83 

Bal.  due  after  Apr.  1,  1860,  ............  $253.50^5. 

NOTE.  —  In  the  following  examples  the  terms  commence  with  the  year,  or  on 
Jan.  1. 

2.  Allowing  interest  monthly  at  6  %  per  annum,  what  sum 
will  be  due  Sept.  1,  1860,  on  the  book  of  a  savings  bank  having 
the  following  entries  ? 

Bay  State  Savings  Institution,  in  account  with  Jane  Ladd. 


Dr. 


Cr. 


1860. 

1860. 

Jan. 

3 

To  cash, 

5 

75 

Jan. 

28 

By  check, 

5 

00 

« 

8 

ii      « 

13 

45 

Feb. 

7 

Cl            tl 

8 

48 

<« 

20 

«      n 

7 

60 

March 

20 

(I      « 

10 

00 

Feb. 

20 

"  check, 

16 

45 

April 

11 

it       il 

12 

76 

« 

27 

"   cash, 

8 

40 

June 

3 

tl       « 

3 

96 

March 

6 

"   check, 

14 

65 

« 

12 

«       (( 

10 

48 

a 

29 

"  cash, 

7 

98 

« 

20 

"  draft, 

17 

48 

April 

25 

ti      >t 

3 

49 

Aug. 

17 

«  check, 

5 

64 

May 

7 

"  draft, 

26 

50 

« 

30 

«(      « 

45 

79 

July 
Aug. 

28 
3 

"  cash, 
"   check, 

15 
18 

68 
45 

t° 

26 

«   cash, 

4 

50 

Ans.  $116.87. 

3.  Interest  at  7  %,  allowed  quarterly,  how  much  was  due  April 
4,  1860,  on  the  following  savings  bank  account  ? 


COMPOUND  INTEREST. 


321 


Dr. 


Detroit  Savings  Institution,  in  account  with  R.  L.  Selden. 


Cr. 


1859. 

1859. 

Jan. 

1 

To  cash, 

47 

50 

May 

12     •     By  check, 

50 

36 

March 

12 

«      >< 

124 

36 

Oct. 

3     j      <•        « 

25 

78 

June 

20 

«      <( 

130 

56 

Nov. 

16           «        « 

36 

48 

Aug. 

3 

«      « 

68 

75 

Dec. 

28     1      "        " 

12  _  50 

1860. 

Jan. 

25 

<;      « 

160 

80 

1                 '             1 

Ans.  $423.22. 

4.  How  much  was  due  Jan.  1,  1860,  on  the  following  account, 
allowing  interest  semi-annually,  at  6  %  per  annum  ? 

Irvings  Savings  Institution,  in  account  with  James  Taylor. 


Dr. 


O. 


1858. 

1858. 

June 

4 

To  cash, 

175 

Sept. 

14 

By  check, 

65 

Nov. 

1 

u   <c 

150 

1859. 

1859. 

July 

25 

«    K 

120 

Feb. 

24 

"  draft, 

200 

Dec.     3 

<l     (( 

80 

Sept. 

10 

«  check, 

56 

1 

Ans.  $337.02. 

5.  Interest  at  5  %,  allowed  according  to  Note  2,  how  much  waa 
due,  Jan.  1,  I860,  on  the  book  of  a  savings  bank  in  the  city  of 
New  York,  having  the  following  entries  ? 


Pr. 


Sixpenny  Savings  Bank,  in  account  with  William  Gallup. 


Cr. 


1858. 

1858. 

" 

Jan. 
March 

i? 

To  check, 

36 
25 

50 

38 

Sept. 
1859. 

16 

By  check, 

36 

10 

Aujr. 

i 

"   cash, 

84 

72 

Jan. 

27 

«      « 

13 

48 

1859. 

March 

1 

«      « 

17 

50 

June 

11 

"   draft, 

50 

00 

Nov. 

16 

"  cash, 

40 

78 

Ans.  §179.10. 

COMPOUND  INTEREST. 

5«?O.   Compound  Interest  is  interest  on  both  principal  and 
interest,  when  the  interest  is  not  paid  when  due. 


NOTE.  —  The  simple  interest  may  be  added  to  the  principal  annually, 
annunlly,  or  quarterly,  as  the  parties  may  agree  ;  but  the  taking  of  compound 
interest  is  not  legal. 

1    What  is  the  compound  interest  of  $640  for  4  years,  af   5 
per  cent.  ? 


822  PERCENTAGE. 

OPERATION. 

$640          Principal  for  1st  year, 
$540  x  1.05  ==  $872                «         "   2d       « 
$672  x  1.05  =  $705.60~         "         "   3d       " 
$705.60  x  1.05  =  $74Q.88~         «         "   4th     « 

$740.88  x  1.05  =  $777.924  Amount     «   4  years, 
640.         Given  principal, 

$137.924    Compound  interest. 

This  illustration  is  sufficient  to  establish  the  following 
RULE.     I.  find  the  amount  of  the  given  principal  at  the  given 
rate  for  one  year,  and  make  it  the  principal  for  the  second  year. 

II.  Find  the  amount  of  this  new  principal,  and  make  it  the 
principal  for  the  third  year,  and  so  continue  to  do  for  the  given 
number  of  years. 

III.  Subtract  the  given  principal  from  the  last  amount ;  the  re- 
mainder will  be  the  compound  interest. 

NOTES. — 1.  When  the  interest  is  paj'able  semi-annually  or  quarterly,  find  the 
amount  of  the  given  principal  for  the  first  interval,  and  make  it  the  principal 
for  the  second  interval,  proceeding  in  all  respects  as  when  the  interest  is  payable 
yearly. 

2.  When  the  time  contains  years,  months,  and  days,  find  the  amount  for  the 
years,  upon  which  compute  the  interest  for  the  months  and  days,  and  add  it  to 
the- last  amount,  before  subtracting. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  compound  interest  of  $750  for  4  years  at  6  per 
cent.?  Ans.  $196.86— 

2.  What  will  $250  amount  to  in  3  years  at  7  per  cent,  compound 
interest?  Ans.  $306.26. 

3.  At  7  per  cent,  interest,  compounded    semi-annually,  what 
debt  will  $1475.50  discharge  in  2J  years?          Ans.  $1752.43. 

4.  Find  the  compound  interest  of  $376  for  3  yr.  8  mo.  15  da., 
at  6  per  cent,  per  annum.  Ans.  $90.84. 

•l»ll.  A  more  expeditious  method  of  computing  compound 
interest  than  the  preceding  is  by  the  use  of  the  compound  interest 
table  on  the  following  page. 


COMPOUND  INTEREST. 


323 


TABLE, 

Showing  the  amount  of  $1,  or  £1,  at  2%,  3,  3£,  4,  5,  6,  7,  and  8  per 
cent.,  compound  interest,  for  any  number  of  years  from  1  to  40. 


Years 

2%perct. 

3  percent. 

S^perct 

4  percent. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

8  per  cent. 

1 

1.025000 

1.030000 

1.035000 

1.040000 

1.050000 

1.060000 

1.070000 

1.080000 

1.050625 

1.060900 

1  071225 

1.081600 

1.102500 

1.123600 

1.144900 

1.166400 

3 

1.076891 

1.092727 

1.108718 

1.124864 

1.157625 

1.191016 

1.225043 

1.259712 

4 

1.103813 

1.125509 

1.147523 

1.169859 

1.215506 

1.262477 

1.310796 

1.360489 

5 

1.131408 

1.159274 

1.187686 

1.216653 

1.276282 

1.338226 

1.402552 

1.469328 

6 

1.159693 

1.194052 

1.229255 

1.265319 

1.340096 

1,418519 

1.500730 

1.586874 

7 

1.188686 

1.229874 

1.272279 

1.315932 

1T407100 

1.503630 

1.605782 

1.713824 

8 

1.218403 

1.266770 

1.316809 

1.368569 

1.477455 

1.593848 

1.718186 

1.850930 

9  1  1.248863 

1.304773 

1.362897 

1.423312 

1.551328 

1.689479 

1.838459 

1.999005 

1J 

1.280085 

1.343916 

1.410599 

1.480244 

1.628885 

1.790848 

1.967151 

2.158925 

11 

1.312087 

1.384234 

1.459970 

1.539454 

1.710339 

1.898299 

2.104852 

2.331639 

12 

1.344889 

1.4257  01 

1.511069 

1.601032 

1.795856 

2.012197 

2.252192 

2.518170 

13 

1.378511 

1.468534 

1.563956 

1.665074 

1.885649 

2.132928 

2.409845 

2.719624 

14 

1.41-2974 

1.512590 

1.618695 

1.731676 

1.979932 

2.260904 

2.578534 

2.937194 

15 

1.44329S 

1.557967 

1.675349 

1.800944 

2.078928 

2.396558 

2.759032 

3.172169 

16 

1.484506 

1.604706 

1.733986 

1.872981 

2.182875 

2.540352 

2.952164 

3.425943 

17 

1.521618 

1.652848 

1.794676 

1.947901 

2.292018 

2.692773 

3.158815 

3.700018 

18 

1.559659 

1.702433 

1.857489 

2.025817 

2.406619 

2.854339 

3.379932 

3.996020 

19 

1.598650 

1.753506 

1.922501 

2.106849 

2.526950 

3.025600 

3.616528 

4.315701 

20 

1.638616 

1.806111 

1.989789 

2.191123 

2.653298 

3.207136 

3.869685 

4.660957 

21 

1.679582 

1.860295 

2.059431 

2.278768 

2.785963 

3.399564 

4.140562 

5.033834 

22 

1.721571 

1.916103 

2.131512 

2.369919 

2.925261 

3.603537 

4.430402 

5.436540 

23 

1.764611 

1.973587 

2.206114 

2.464716 

3.071524 

3.819750 

4.740530 

5.871464 

24 

1.808726- 

2.032794 

2.283328 

2.563304 

3.225100 

4.048935 

5.072367 

6.341181 

25 

1.853944 

2.093778 

2.363245 

2.665836 

3.386355 

4.291871 

5.427433 

6.848475 

26 

1.900293 

2.156591 

2.445959 

2.772470 

3.555673 

4.549383 

5.807353 

7.396353 

27 

1.947800 

2.221289 

2.531567 

2.883369. 

3.733456 

4.822346 

6.213868 

7.988062 

28 

1.996495 

2.287928 

2.620172 

2.998703 

3.920129 

5.111687 

6.648838 

8.627106 

29 

2.046407 

2.356566 

2.711878 

3.118651 

4.116136 

5.418388 

7.114257 

9.317275 

30 

2.097568 

2.427262 

2.806794 

3.243398 

4.321942 

5.743491 

7..612255 

10.062657 

31  1  2.150007 

2.500080 

2.905031 

3.373133 

4.538040 

6.088101 

8.145113 

10.867669 

32  1  2.203757 

2.575083 

3.006708 

3.508059 

4.764942 

6.4533S7 

8.715271 

11.737083 

33 

2.258851 

2.652335 

3.111942 

3.648381 

5.003189 

6.840590 

9.325340 

12.676050 

34 

2.315322 

2.731905 

3.220860 

3.794316 

5.253348 

7.251025 

9.978114 

13.690134 

35 

2.373205 

2.813862 

3.333590 

3.946089 

5.516015 

7.68C087 

10.676582 

14.785344 

36 

2.432535 

2.896278 

3.450266 

4.103933 

5.791816 

8.147252 

11.423942 

15.968172 

37 

2.493349   2.985227 

3.571025 

4.268090 

6.0S1407 

8.636087 

12.223618 

17.245626 

38 

2.555682  3.074783 

3.696011 

4.438813 

6.385477 

9.154252 

13.079271 

18.625276 

39 

2.619574  i  3.167027 

3.825372 

4.616366 

6.704751 

9.703508 

13.994820 

20.115298 

40 

2.685064  |  3.26203S 

3.959260 

4.801021 

7.039989 

10.285718 

14.974458 

21.724522 

324  PERCENTAGE. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  amount  of  $300  for  4  years  at  6  per  cent,  com- 
pound interest,payable  semi-annually  ? 

OPERATION.  ANALYSIS.    The  amount  of  $1  at  6  per  cent., 

$1  26677  compound  interest  payable  semi-annually,  is 

300  the  same  as  the  amount  of  $1  at  3  per  cent., 

AOOA'AO-IQA  compound   interest    payable    annually.     We 

therefore  take,  from  the  table,  the  amount  of 

$1  for  8  years  at  3  per  cent.,  and  multiply  this  amount  by  the  given 
principal. 

2.  What  is  the  amount  of  $536.75  for  12  yr.  at  8  per  cent,  com- 
pound interest  ?  Ans.  $1351.63. 

3.  What  sum  placed  at  simple  interest  for  2  yr.  9  mo.  12  da., 
at  7  per  cent.,  will  amount  to  the  same  as  $1275,  placed  at  com- 
pound interest  for  the  same  time   and  at  the  same  rate,  payable 
semi-annually?  Ans.  $1292.51  —  . 

4.  At  8  per  cent,  interest  compounded  quarterly,  how  much 
will  $1840  amount  to  in  1  yr.  10  mo.  20  da.  ?     Ans  $2137.06. 

5.  A  father  at  his  death  left  $15000  for  the  benefit  of  his  only 
son,  who  was  12  yr.  7  mo.  12  da.  old  when  the  money  was  de- 
posited ;  the  same  was  to  be  paid  to  him  when  he  should  be  21 
years  of  age,  together  with  7  per  cent,  interest  compounded  semi- 
annually.     How  much  was  the  amount  paid  him  ? 

6.  What  sum  of  money  will  amount  to  $2902.263  in  20  years, 
at  7  %  compound  interest?  Ans.  $750. 


PROBLEMS  IN  INTEREST. 
PROBLEM   I. 

.   Given,  the  time,  rate  per  cent.,  and  interest,  to 
find  the  principal. 

1.  Y/hat  sum  of  money  will  gain  $87.42  in  4  years,  at  6  per 
cent.  ? 

OPERATION.  ANALYSIS.     Since   $.24 

$.24,  interest  of  $1  for  4  years.         is  the  interest  of  $l  for  4 
$87.42  -r-  .24  .  $364.25,  Ans.         years  at  6  per  cent.,  $87.42 

must  be  the  interest  of  as 


PROBLEMS  IN  INTEREST.  325 

many  dollars,  for  the  same  time  and  at  the  same  rate,  as  $.24  is  con- 
tained times  in  $87.42.  Dividing,  we  obtain  $364.25,  the  required 
principal.  Hence  the 

RULE.     Divide  the  given  interest  by  the  interest  of  $1  for  the 
given  time  at  the  given  rate. 

EXAMPLES    FOR   PRACTICE. 

1.  What  sum  of  money,  invested  at  6£  per  cent.,  will  produce 
$279.825  in  1  yr.  6  mo.?  Am.  $2870. 

2.  What  sum  will  produce  $63.75  interest  in  6  mo.  24  da.  at 
7  £  per  cent.  ? 

3.  What  sum  will  produce  $12 £  interest  in  10  days  at  10  per 
cent.?  Am.  $4500. 

4.  What  sum  must  be  invested  in  real  estate  paying  12  ^  per 
cent,  profit  in  rents,  to  give  an  income  of  $3125  ? 

5.  What  is  the  value  of  a  house  and  lot  that  pays  a  profit  of  9| 
per  cent,  by  renting  it  at  $30  per  month  ? 

6.  What  sum  of  money,  put  at  interest  6  yr.  5  mo.  11  da.,  at 
7  per  cent.,  will  gain  $3159.14  ?  Am.  $7000, 

7.  What  sum  of  money  will  produce  $69.67  in  2  yr.  9  mo.  at 
6  %  compound  interest  ?  Am.  $400. 

8.  What   principal    at   6   %   compound    interest  will  produce 
$124.1624  in  1  yr.  6  mo.  15  da.  ?  Am.  $1314.583. 

PROBLEM    II. 

553.   Given,  the  time,  rate  per  cent,  and  amount,  to 
find  the  principal. 

1.  What  sum  of  money  in  2  years  6  months,  at  7  per  cent., 
will  amount  to  $136.535  ? 

OPERATION.  ANALYSIS.    Since 

$1.175,  amount  of  $1  for  2  yr.  6  mo.          $L175  i^  the  amount 

$136.535  -  1.175  =  $116.20,  Am.  of  $1  for  2  ^ar8  6 

months,    at    7    per 

cent.,  $136.535  must  be  the  amount  of  as  many  dollars,  for  the  same 
time  and  at  the  same  rate,  as  $1.175  is  contained  times  in  $136.535- 
Dividing,  we  obtain  $116.20,  the  required  principal.     Hence  the 
28 


326  PERCENTAGE. 

RULE.     Divide  the  given  amount  by  the  amount  of  $1  for  the 
given  time  at  the  given  rate. 

EXAMPLES    FOR   PRACTICE. 

1.  What  principal  in  2  yr.  3  mo.  10  da.,  at  5  per  cent.,  will 
amount  to  $1893  61$  ?  Ans.  $1700. 

2.  A  note  which  had  run  3  yr.   5  mo.    12  da.  amounted  to 
$681.448,  at  6  per  cent. ;  how  much  was  the  face  of  the  note  ? 

3.  What  sum  put  at  interest  at  3^  per  cent.,  for  10  yr.  2  mo., 
will  amount  to  $15660? 

4.  What  is  the  interest  of  that  sum  for  2  yr.  8  mo.  29  da.,  at  7 
per    cent.,  which   at   the    same   time   and  rate,   will   amount  to 
$1568.97?  Ans.  $253.057. 

5.  What  is  the  interest  of  that  sum  for  243  days  at  8  per  cent., 
which  at  the  same  time  and  rate,  will  amount  to  $11119.70  ? 

6.  What  principal  in  4  years  at  6  per  cent,  compound  interest, 
will  amount  to  $8644.62  ?  Ans.  $6847.34. 

7.  What  sum  put  at  compound  interest  will  amount  to  $26772.96, 
in  10  yr.  5  mo.,  at  6  per  cent.  ? 

Ans.  $14585.24. 

PROBLEM   III. 

554.   Given,  the  principal,  time,  and  interest,  to  find 
the  rate  per  cent. 

1.    I  received   $315   for  3   years'  interest  on  a  mortgage  of 
$1500 ;  what  was  the  rate  per  cent.  ? 

OPERATION.  ANALYSIS.      Sines 

$15  00  $45  is  the  interest  on 

3  the    mortgage    for    3 

years   at  1  per  cent., 
$45.00,  int.  for  3  yr.  at  1  %.          fa  must  £  the  ^ 

$315  -f-  $45  =a  7  %,  Ans.  terest  on  the  mortgage 

for  the  same  time,  at 

as  many  times  1  per  cent,  as  $45  is  contained  times  in  $315.     Divid- 
ing, and  we  obtain  7,  the  required  rate  per  cent.     Hence  the 

RULE.     Divide  the  given  interest  by  the  interest  on  the  principal 
for  the  given  time  at  1  per  cent. 


PROBLEMS  IN  INTEREST.  327 

EXAMPLES   FOR   PRACTICE. 

1.  If  I  loan  $750  at  simple  interest,  and  at  the  end  of  1  yr.  3 
tno.  receive  $796.871,  what  is  the  rate  per  cent.  ?  Ans.  5. 

2  If  I  pay  $10.58  for  the  use  of  $1700,  28  days,  what  is  the 
rate  of  interest?  Ans.  8-fper  cent. 

3.  Borrowed  $600,  and  at  the  end  of  9  yr.  6  mo.  returned 
$856.50  ;  what  was  the  rate  per  cent.  ? 

4.  A  man  invests  17266.28,  which  gives  him  an  annual  income 
of  $744.7937;  what  rate  of  interest  does  he  receive  ? 

5.  If  C  buys  stock  at  30  per  cent,  discount,  and  every  6  months 
receives  a  dividend  of  4  per  cent.,  what  annual   rate  of  interest 
does  he  receive  ?  Ans.  11|  per  cent. 

6.  At  what  rate  per  annum  of  simple  interest  will  any  sum  of 
money  double  itself  in  4,  6,  8,  and  10  years,  respectively  ? 

7.  At  what  rate  per  annum  of  simple  interest  will  any  sum 
triple  itself  in  2,  5,  7,  12,  and  20  years,  respectively  ? 

8.  A  house  that  rents  for  $760.50  per  annum,  cost  $7800  :  what 
%  does  it  pay  on  the  investment?  Ans.  9|  per  cent. 

9.  I  invest  $35680  in  a  business  that  pays  me  a  profit  of  $223  a 
month  ;  what  annual  rate  of  interest  do  I  receive  ?      Ans.  7  \  %. 

PROBLEM    IV. 

555.  Given,  the  principal,  interest,  and  rate,  to  find 
the  time. 

1.  In  what  time  will  $924  gain  $151.536,  at  6  per  cent.? 

OPERATION.  ANALYSIS.      Since 

$924  $55.44  is  the  interest 

.06  of  $924  for  1  year  at 

.,    $151,536 


Arr    AA       '      i.         n<&AO^     f         1  L   r>    ^ 

$55.44,  int.  of  $924  for  1  yr.  at  6  %. 
$151.536  -  $55.44  =  2.73  of  the  game 

2.73  yr.  =  2  yr.  8  mo.  24  da.,  Ans.         the    same    rate    per 

cent.,    for    as    many 

years  as  $55.44  is  contained  times  in  $151.536,  which  is  2.73  times. 
Reducing  the  mixed  decimal  to  its  equivalent  compound  number, 
we  have  2  years  8  months  24  days,  the  required  time.  Hence  the 


828  PERCENTAGE. 

RULE.  Divide  the  given  interest  by  the  interest  on  the  principal 
for  1  year ;  the  quotient  will  be  the  required  time  in  years  and 
decimals.  '« 

EXAMPLES    FOR   PRACTICE. 

1.  In  what  time  will  $273.51  amount  to  $312.864,  at  7  per 
cent.  ?  Ans.  2  yr.  20  da. 

2.  How  long  must  $650.82  be  on  interest  to  amount  to  $761.44, 
at  5  per  cent.  ?  Ans.  3  yr  4  mo.  24  da. 

3.  How  long  will  it  take  any  sum  of  money  to  double  itself  by 
simple  interest  at  3,  4J,  6,  7,  and  10  per  cent.  ?     How  long  to 
quadruple  itself?      ^  j  To  double  itself  at  3  %,  33£  yr 

S'  (  To  quadruple  itself  at  3  %.  100  yr. 

4.  In  what  time  will  $9750  produce  $780  interest,  at  2  per 
cent,  a  month  ? 

5.  In  what  time  will  $1000  draw  $1171.353  at  6  per  cent,  com, 
pound  interest  ? 

ANALYSIS.  $1171.353-r-1000=$1.171353,  the  amount  of  $1  for  the 
required  time.  From  the  table,  $1,  in  2  years,  will  amount  to  $1.1236 ; 
hence  $1.171353 — $1.1236— $.047753,  the  interest  which  must  accrue 
on  $1.1236  for  the  fraction  of  a  year;  and  $1.1236  X  .06  =  $.067416 ; 
$.047753  -7-  $.067416  =  .7083  yr.  =  8  mo.  15  da. 

Ans.  2  yr.  8  mo.  15  da. 

6.  In  what  time  will  $333  amount  to  $376.76  at  5  per  cent 
compound  interest,  payable  semi-annually  ? 

7.  In  what  time  will  any  sum  double  itself  at  6  %  compound 
interest  ?     At  7  %  ?  Ans.  to  last,  10  yr.  2  mo.  26  da. 


DISCOUNT. 

•156.  Discount  is  an  abatement  or  allowance  made  for  the 
payment  of  a  debt  before  it  is  due 

5,17.  The  Present  Worth  of  a  debt,  payable  at  a  future  time 
without  interest,  is  such  a  sum  as,  being  put  at  legal  interest,  will 
amount  to  the  given  debt  when  it  become?  due. 

1.  What  is  the  present  worth  and  what  the  discount  of  $642.12 
to  be  paid  4  yr.  9  mo.  27  da.  hence,  money  being  worth  7  per 
cent.  ? 


DISCOUNT.  329 

OPERATION.  ANALYSIS.    Since  $1  is  the. 

$1.33775,  Amount  of  81.  present  worth   of  $1.33775 

$642.12  -T-  1.33775  =  $480  for   the  given  time   at  the 

$642. 12,  given  sum..  given  rate  of  interest,  the 

480.         present  worth,  present    worth    of   $642.12 

$162 \.12,  discount.  must }™  as  many  dollars  as 

$1.33775  is  contained  times 

in  $642.12.  Dividing,  and  we  obtain  $480  for  the  present  worth,  and 
subtracting  this  sum  from  the  given  sum,  we  have  $162.12,  the  dis- 
count. Hence  the  following 

RULE.  I  Divide  the  given  sum  or  debt  by  the  amount  of  $1 
for  the  given  rate  and  time;  the  quotient  will  be  the  present  worth 
of  the  debt. 

II.  Subtract  the  present  worth  from  the  given  sum  or  debt ;  the 
remainder  will  be  the  discount. 

NOTES.  —  1.  The  terms  present  worth,  discount,  and.  debt,  are  equivalent  to 
principal,  interest,  and  amount.  Hence,  when  the  time,  rate  per  cent.,  and 
amount  are  given,  the  principal  maybe  found  by  Prob.  II,  (553);  and  the 
interest  by  subtracting  the  principal  from  the  amount. 

2.  When  payments  are  to  be  made  at  different  times  without  interest,  find  the 
present  worth  of  each  payment  separately.  Their  sum  will  be  the  present  worth 
of  the  several  payments,  and  this  sum  subtracted  from  the  sum  of  the  several 
payments  will  leave  the  total  discount. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  present  worth  of  a  debt  of  $385. 31|,  to  be  paid 
in  5  mo.  15  da.,  at  6  %  ?  Ans.  $375. 

2.  How  much  should  be  discounted  for  the  present  payment  of 
a  note  for  $429.986,  due  in  1  yr.  6  mo.  1  da.,  money  being  worth 
5i  %  ?  Ans.  $32.826. 

3.  Bought  a  farm  for  $2964.12  ready  money,  and  sold  it  again 
for  $3665.20,  payable  in  1  yr.  6  mo.    How  much  would  be  gained 
in  ready  money,  discounting  at  the  rate  of  8  %  ? 

4.  A  man  bought  a  flouring  mill  for  $25000  cash,  or  for  $12000 
payable  in  6  mo.  and  $15000  payable  in  1  yr.  3  mo.    He  accepted 
the  latter  offer ;  did  he  gain  or  lose,  and  how  much,  money  being 
worth  to  him  10  per  cent.  ?  An*.  Gained  $238.10. 

5.  B  bought  a  house  and  lot  April  1,  1860,  for  which  he  waa 
to  pay  $1470  on  the  fourth  day  of  the  following  September,  and 

28* 


g3Q  PERCENTAGE. 

$2816.80  Jan  1,  1861.  If  he  could  get  a  discount  of  10  per 
cent,  for  present  payment,  how  much  would  he  gain  by  borrowing 
the  sum  at  7  per  cent.,  and  how  much  must  he  borrow? 

6.  What  is  the  difference  between  the  interest  and  the  discount 
of  $576,  due  1  yr.  4  mo.  hence,  at  6  per  cent.  ? 

7.  A  merchant  holds  two  notes  against  a  customer,  one  for 
$243.16,  due  May  6,  1861,  and  the  other  for  $178.64,  due  Sept. 
25,  1861 ;  how  much  ready  money  would  cancel  both  the  notes 
Oct.  11,  I860,  discounting  at  the  rate  of  7  %  ?    Am.  $401.29—. 

8.  A  speculator  bought  120  bales  of  cotton,  each  bale  containing 
488  pounds,  at  9  cents  a  pound,  on  a  credit  of  9  months  for  the 
amount.     He  immediately  sold  the  cotton  for  $6441.60  cash,  and 
paid  the  debt  at  8  %  discount ;  how  much  did  he  gain  ? 

9.  Which  is  the  more  advantageous,  to  buy  flour  at  $6.25  a 
barrel  on  6  months,  or  at  $6.50  a  barrel  on  9  months,  money  being 
worth  8  %  ? 

10.  How  much  may  be  gained  by  hiring  money  at  5  %  to  pay 
a  debt  of  $6400,  due  8  months  hence,  allowing  the  present  worth 
of  this  debt  to  be  reckoned  by  deducting  5   %  per  annum  dis- 
count? Ans.  $7.11£. 

BANKING. 

558.  A  Bank  is  a  corporation  chartered  by  law  for  the  pur- 
pose of  receiving  and  loaning  money,  and  furnishing  a  paper 
circulation. 

559.  A  Promissory  Note  is  a  written  or  printed  engagement 
to  pay  a  certain  sum  either  on  demand  or  at  a  specified  time. 

5OO.  Bank  Notes,  or  Bank  Bills,  are  the  notes  made  and 
issued  by  banks  to  circulate  as  money.  They  are  payable  in  specie 
at  the  banks. 

NOTE. — A  bank  which  issues  notes  to  circulate  as  money  is  called  a  lank  of 
issue  ;  one  which  lends  money,  a  bank  of  discount ;  and  one  which  takes  charge 
of  money  belonging  to  other  parties,  a  bank  of  deposit.  Some  banks  perform 
two  and  some  all  of  these  duties. 

561.  The  Maker  or  Drawer  of  a  note  is  the  person  by  whom 
the  note  is  signed ; 

562.  The  Payee  is  the  person  to  whose  order  the  note  is  made 
payable;  and 


BANKING.  33] 

563.   The  'Holder  is  the  owner. 

564L  A  Negotiable  Note  is  one  which  may  be  bought  and 
sold,  or  negotiated.  It  is  made  payable  to  the  bearer  or  to  the 
order  of  the  payee. 

565c  Indorsing  a  note  by  a,  payee  or  holder  is  the  act  of 
writing  his  name  on  its  back. 

NOTKS. — 1.  If  a  note  is  payable  to  the  bearer,  it  may  be  negotiated  without 
indorsement. 

2.  An  indorsement  makes  the  indorser  liable  for  the  payment  of  a  note,  if  the 
maker  fails  to  pay  it  when  it  is  due. 

3.  A  note  should  contain  the  words  "value  received,"  and  the  sum  for  which 
it  is  given  should  be  written  out  in  words. 

566.  The  Face  of  a  note  is  the  sum  made  payable  by  the 
note. 

567.  Days  of  Grace  are  the  three  days  usually  allowed  by 
law  for  the  payment  of  a  note  after  the  expiration  of  the  time 
specified  in  the  note. 

568  The  Maturity  of  a  note  is  the  expiration  of  'the  days 
of  grace ;  a  note  is  due  at  maturity. 

NOTE. — No  grace  is  allowed  on  notes  payable  "  on  demand,"  without  grace. 
In  some  States  no  grace  is  allowed  on  notes,  and  their  maturity  is  the  expira- 
tion of  the  time  mentioned  in  them. 

569.  Notes  may  contain  a  promise  of  interest,  which   will 
be  reckoned  from  the  date  of  the  note,  unless  some  other  time  be 
specified. 

NOTE. — A  note  is  on  interest  from  the  day  it  is'due,  even  though  no  mention 
be  made  of  interest  in  the  note. 

570.  A  Notary,  or  Notary-Public,  is  an  officer  authorized 
by  law  to  attest  documents  or  writings  of  any  kind  to  make  them 
authentic. 

571.  A  Protest  is  a  formal  declaration  in  writing,  made  by  a 
Notary-Public,  at  the  request  of  the  holder  of  a  note,  notifying 
the  maker  and  the  indorsers  of  its  non-payment. 

NOTES. — 1.  The  failure  to  protest  a  note  on  the  third  day  of  grace  releases  the  iii- 
dorsers  from  all  obligation  to  pay  it. 

2.  If  the  third  day  of  grace  or  the  maturity  of  a  note  occurs  on  Sunday  or  a  legal 
holiday,  it  must  be  paid  on  the  day  previous. 

572.  Bank  Discount  is  an  allowance  made  to  a  bank  for  the 
payment  of  a  note  before  it  becomes  due. 


332  PERCENTAGE. 

•17JI.   The  Proceeds  of  a  note  is  the  sum  received  for  it 
discounted,  and  is  equal  to  the  face  of  the  note  less  the  discount. 

«574,  The  transaction  of  borrowing  money  at  banks  is  con- 
ducted in  accordance  with  the  following  custom  :  The  borrower 
presents  a  note,  either  made  or  indorsed  by  himself,  payable  at  a 
specified  time,  and  receives  for  it  a  sum  equal  to  the  face  ;  less 
the  interest  for  the  time  the  note  has  to  run.  The  amount  thus 
withheld  by  the  bank  is  in  consideration  of  advancing  money  on 
the  note  prior  to  its  maturity. 

NOTES.  —  1.  A  note  for  discount  at  bank  must  be  made  payable  to  the  order 
of  some  person,  by  whom  it  must  be  indorsed. 

2.  The  business  of  buying  or  discounting  notes  is  chiefly  carried  on  by  banks 
and  brokers. 

The  law  of  custom  at  banks  makes  the  bank  discount 


of  a  note  equal  to  the  simple  interest  at  the  legal  rate,  for  the 
time  specified  in  the  note.  As  the  bank  always  takes  the  interest 
at  the  time  of  discounting  a  note,  bank  discount  is  equal  to  simple 
interest  paid  in  advance.  Thus,  the  true  discount  of  a  note  for 
$153,  which  matures  in  4  months  at  6  %,  is  $153—  'fsoo  = 
$3.00,  and  the  bank  discount  is  $153  x  .02  =»  $3.06.  Since  the 
interest  of  $3,  the  true  discount,  for  4  months  is  $3  x  .02  =  $.06, 
we  observe  that  the  bank  discount  of  any  sum  for  a  given  time  is 
greater  than  true  discount,  by  the  interest  on  the  true  discount 
for  the  same  time. 

NOTE.  —  Many  banks  take  only  true  discount. 

CASE   I. 

576.  Given,  the  face  of  a  note,  to  find  the  discount 
and  the  proceeds. 

RULE.  I.  Compute  the  interest  on  the  face  of  the  note  for  three 
days  more  than  the  specified  time;  the  result  will  be  the  discount. 

II.  Subtract  the  discount  from  the  face  of  the  note;  the  re- 
mainder will  be  the  proceeds. 

NOTES.  —  1.  When  a  note  is  on  interest,  payable  at  a  future  specified  time,  the 
amount  is  the  face  of  the  note,  or  the  sum  made  payable,  and  must  be  made  the 
basis  of  discount. 

2.  To  indicate  the  maturity  of  a  note  or  draYt,  a  vertical  line  (  |  )  is  used,  with 
the  day  at  which  the  note  is  nominally  due  on  the  left,  and  the  date  of  maturity 
on  the  right;  thus,  Jan.  7  |  ,  0. 


BANKING.  333 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  bank  discount,  and  what  are  the  proceeds  of  a 
note  for  $1487  due  in  30  days  at  6  per  cent.  ? 

Ans.   Discount,  $8.18;  Proceeds,  $1478.82. 

2.  What  are  the  proceeds  of  a  note  for  $384.50  at  90  days,  if 
discounted  at  the  New  York  Bank? 

3.  Wishing  to  borrow  $1000  of  a  Southern  bank  that  is  dis- 
counting paper  at  8  per  cent.,  I  give  my  note  for  $975,  payable 
in  60  days ;  how  much  more  will  make  up  the  required  amount  ? 

4.  A  man  sold  his  farm  containing  195  A.  2  R.  25  P.  for  $27.50 
an  acre,  and  took  a  note  payable  in  4  mo.  15  da.  at  7  %  interest. 
Wishing  the  money  for  immediate  use,  he  got  the  note  discounted 
at  a  bank;  how  much  did  he  receive  ?  Ans.  $5236.169. 

5.  Find  the  day  of  maturity,  the  term  of  discount,  and  the  pro- 
ceeds of  the  following  notes : 

$1962-^%.  DETROIT,  July  26,  1860. 

Four  months  after  date  I  promise  to  pay  to  the  order  of  James 
Gillis  one  thousand  nine  hundred  sixty-two  and  T4^  dollars  at  the 
Exchange  Bank,  for  value  received.  JOHN  DEMAREST. 

Discounted  Aug.  26,  at  7%. 

Ans.  Due  Nov.  26  |  29;  term  of  discount  95  days;  proceeds; 
$1926.20. 

%.  BALTIMORE,  April  19,  1859. 


6.  Ninety  days  after  date  we  promise  to  pay  to  the  order  of 
King  &  Dodge  one  thousand  sixty-six  and  y7^  dollars  at  the  Citi. 
zens'  Bank,  for  value  received.  CASE  &  SONS. 

Discounted  May  8,  at  6  % . 

Ans.  Due  July  1 8  |  f , ;  term  of  discount,  74  da. ;   proceeds, 
$1053.59. 
$784/0%.  MOBILE,  June  20,  1861. 

7.  Two  months  after  date  for  value  received  I  promise  to  pay 
George  Thatcher  or  order  seven  hundred  eighty-four  and  T7^  dol- 
lars at  the  Traders'  Bank.  WM.  HAMILTON. 

Discounted  July  5,  at  8  %. 


334  PERCENTAGE. 


$1845^.  CHICAGO,  Jan.  31,  1862. 

8.  One  month  after  date  we  jointly  and  severally  agree  to  pay 
to  W.  H.  Willis,  or  order,  one  thousand  eight  hundred  forty  7five 
and  -f^Q  dollars  at  the  Marine  Bank. 

FAYSON  &  WILLIAMS. 

Discounted  Jan.  31,  at  2  %  a  month. 

Ans.  Due  Feb.  28  |  March  3;  term  of  discount,  31  da.;  pro- 
ceeds, $1807.36. 

9.  What  is  the  difference  between  -the  true  and  the  bank  dis- 
count of  $950,  for  3  months  at  7  per  cent.  ?  Ans.  $.29. 

10.  What  is  the  difference  between  the  true  and  the  bank  dis- 
count of  $1375.50,  for  60  days  at  6  per  cent.  ? 

CASE    II. 

577.   Given,  the  proceeds  of  a  note,  to  find  the  face. 
1.  For  what  sum  must  I  draw  my  note  at  4  months,  interest 
6  %,  that  the  proceeds  when  discounted  in  bank  shall  be  $750  ? 

OPERATION.  ANALYSIS.       We 

$1.0000  first  obtain  the  pro- 

.0205,  disc't  on  $1  for  4  mo.  3  da.  ceeds  of  $1  by  the 

(T9795,  proceeds  of  $1.  last  case;  then,  since 

$750  nh  .9795  =  $765.696,  Ans.  ^9795  1S  the  Pr?- 

ceeds  of  $1,  $750  is 

the  proceeds  of  as  many  dollars  as  $.9795  is  contained  times  in  $750. 
Dividing,  we  obtain  the  required  result.     Hence  the 

RULE.  Divide  the  proceeds  l>y  the  proceeds  of  $lybr  the  time 
and  rate  mentioned  ;  the  quotient  will  be  the  face  of  the  note. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  face  of  a  note  at  60  days,  the  proceeds  of  which, 
when  discounted  at  bank  at  6  %,  are  $1275?    Ans.  $1288.53. 

2.  If  a  merchant  wishes  to  draw  $5000  at  bank,  for  what  sum 
must  he  give  his  note  at  90  days,  discounting  at  6  per  cent.  ? 

Ans.  $5078.72. 

3.  The  avails  of  a  note  having  3  months  to  run,  discounted  at 
a  bank  at  7  %,  were  $276.84;  what  was  the  face  of  the  note  ? 


BANKING.  335 

4.  James  T.  Fisher  buys  a  bill  of  merchandise  in  New  York  at 
cash  price,  to  the  amount  of  $1486.90,  and  gives  in  payment  his 
note  at  4  months  at  7^  % ;  what  must  be  the  face  of  the  note  ? 

5.  Find  the  face  of  a  6  mo.  note,  the  proceeds  of  which,  dis- 
counted at  2  %  a  month,  are  $496.  AUK.  $564.92. 

6.  For  what  sum  must  a  note  be  drawn  at  30  days,  to  net 
$1200  when  discounted  at  5  %  ? 

7.  Owing  a  man  $575,  I  give  him  a  60  day  note;  what  should 
be  the  face  of  the  note,  to  pay  him  the  exact  debt,  if  discounted 
at  1^  %  a  month?  Ans.  $593.70. 

8.  What  must  be  the  face  of  a  note  which,  when  discounted  at 
a  broker's  for  110  days  at  1  %  a  month,  shall  give  as  its  proceeds 
$187.50? 

CASE   III. 

578.   Given,  the  rate  of  bank  discount,  to  find  the 
corresponding  rate  of  interest. 

I.  A  broker  discounts  30  day  notes  at  1^  ^  a  month;  what 
rate  of  interest  does  his  money  earn  him  ? 

OPERATION.  ANALYSIS.  If  we  assume 

30  day  notes  =  33  days'  time.  $1°°  as   the  face   of   the 

$100,         base.  note,  the   discount  for  33 

1.65,  discount  for  33  days.  days  at  1^  a  month  will 

$98.35,  proceeds.  be  $L65  and  the  Proceeds 

$1,65  -f-. 0901541  =  18-fl ?ff4,  %,  Ans.    $98-35-      We    then    have 

$98.35  principal,  $1.65  in- 
terest, and  33  days  time,  to  find  the  rate  per  cent,  per  annum,  which 
we  do  by  (554).  Hence  the 

KULE.     I.  Find  the  discount  and  the  proceeds  of  $1  or  $100 
for  the  time  the  note  has  to  run. 

II.  Divide  the  discount  by  the  interest  of  the  proceeds  at  1  per 
cent,  for  the  same  time. 

EXAMPLES    FOR    PRACTICE. 

1.  What  rate  of  interest  is  paid,  when  a  note  payable  in  30 
days  is  discounted  at  6  per  cent.  ?  Ans. 


836  PERCENTAGE. 

2.  A  note  payable  in  2  months  is  discounted  at  2  %  a  month; 
what  rate  of  interest  is  paid?  Ans.  25ff-g  <J0. 

3,,  When  a  note  payable  in  90  days  is  discounted  at  \\  v/0  a 
month,  what  rate  of  interest  is  paid?  Ans.  18}f£y  %. 

4.  What  rate  of  interest  corresponds  to  5,  6,  7,  10,  12  %  dis- 
count on  a  note  running  10  months  without  grace  ? 

5.  What  rate  of  interest  does  a  man  pay  who  has  a  60  day 
note  discounted  at  |,  1,  2,  2^,  3  %  a  month? 

CASE   IV. 

579.  Given,  ih&  rate  of  interest,  to  find  the  corres- 
ponding rate  of  bank  discount. 

1    A  broker  buys  60  day  notes  at  such  a  discount  that  his 

money  earns  him  2  %  a  month ;  what  is  his  rate  °fc  of  discount  ? 

OPERATION.  ANALYSIS.     If  we  assume 

60  da.  -f  3  da,  =  63  da.  $100  as  the  proceeds  of  a 

$100          base.  note,  the  interest  for  63  days 

4.20,  interest  for  63  da.  at  24  per  cent,  will  be  $4.20, 

$104.20,  amount       «       «  and  the  amount  or  f™e  of 

$4.20  -*-  .18235  =  23^V7-  %,  Ans     the  note  will  be  $104.20.  We 

then  have  $104.20  the  prin- 
cipal, $4.20  the  interest,  and  63  days  the  time,  to  find  the  rate  per 
cent.,  which  we  do  by  (554)  as  in  the  last  case.  Hence  the 

RULE.  I.  Find  the  interest  and  the  amount  o/$l  or  $100  for 
the  time  the  note  lias  to  run. 

II.  Divide  the  interest  by  the  interest  on  the  amount  at  1  per 
cent,  for  the  same  time. 

EXAMPLES    FOR   PRACTICE. 

1.  What  rates  of  bank  discount  on  30  day  notes  correspond  to 
5,  6,  7,  10  per  cent,  interest  ? 

2.  At  what  rate  should  a  3  months'  note  be  discounted  to  pro- 
duce 8  %  interest?  Ans.  7  Iff?   %. 

3.  At  what  rates  should  60  day  notes  be  discounted  to  pay  to  a 
broker  1,  U,  2,  2£  %  a  month? 

4.  At  what  rate  must  a  note  payable  18  months  hence,  without 
grace,  be  discounted  to  produce  7  %  interest?     Ans. 


EXCHANGE.  337 

EXCHANGE. 

58O.  Exchange  is  a  method  of  remitting  money  from  one 
place  to  another,  or  of  making  payments  by  written  orders. 

081.  A  Bill  of  Exchange  is  a  written  request  or  order  upon 
one  person  to  pay  a  certain  sum  to  another  person,  or  to  his  order, 
at  a  specified  time. 

582.  A  Sight  Draft  or  Bill  is  one  requiring  payment  to  be 
made  "at  sight,"  which  means,  at  the  time  of  its  presentation  to 
the  person  ordered  to  pay  In  other  bills,  the  time  specified  is 
usually  a  certain  number  of  days  "  after  sight." 

There  are  always  three  parties  to  a  transaction  in  exchange,  and 
usually  four : 

588.  The  Drawer  or  Maker  Is  the  person  who  signs  the 
order  or  bill ; 

58  4U  The  Drawee  is  the  person  to  whom  the  order  is  ad- 
dressed ; 

585.  The  Payee  is  the  person  to  whom  the  money  is  ordered 
to  be  paid  ;  and 

586.  The  Buyer  or  Remitter  is  the  person  who  purchases 
the  bill.     He  may  be  himself  the  payee,  or  the  bill  may  be  drawn 
in  favor  of  any  other  person. 

587.  The  Indorsement  of  a  bill  is  the  writing  upon  its  back, 
by  which  the  payee  relinquishes  his  title,  and   transfers  the  pay- 
ment to  another.     The  payee  may  indorse  in  blank  by  writing  his 
name  only,  which  makes  the  bill  payable  to  the  bearer,  and  con- 
sequently transferable  like  a  bank  note;  or  he  may  accompany  his 
signature  by  a  special  order  to  pay  to  another  person,  who  in  his 
turn  may  transfer  the  title  in  like  manner.     Inclorsers  become  sep- 
arately responsible  for  the  amount  of  the  bill,  in  case  the  drawee 
fails  to  make  payment.      A  bill  made  payable  to  the  bearer  is 
transferable  without  indorsement. 

588.  The   Acceptance  of  a  bill  is  the   promise  which  the 
drawee  makes  when  the  bill  is  presented  to  him  to  pay  it  at  ma- 
turity; this   obligation    is   usually  acknowledged  by  writing  the 
word  "  Accepted,"  with  his  signature,  across  the  face  of  the  bill, 

29  w 


338  PERCENTAGE. 

NOTES. — 1.  In  this  country,  and  in  Great  Britain,  three  days  of  grace  are  al- 
lowed fur  the  payment  of  a  bill  of  exchange,  after  the  time  specified  has  expired. 
In  regard  to  grace  on  isiyltt  bilh,  however,  custom  is  variable  ;  in  New  York. 
Pennsylvania,  Virginia,  and  some  other  States,  no  grace  is  allowed  on  sight  bills, 

2.  When  a  bill  is  protested  for  non-acceptance,  the  drawer  is  obligated  to  pay 
it  immediately,  even  though  the  specified  time  has  not  expired. 

Exchange  is  of  two  kinds  —  Domestic  and  Foreign. 

089.  Domestic  or  Inland  Exchange  relates  to  remittances 
made  between  different  places  of  the  same  country. 

NOTE. — An  Inland  Bill  of  Exchange  is  commonly  called  a  Draft. 

090.  Foreign  Exchange  relates  to  remittances  made  between 
different  countries. 

091.  A  Set  of  Exchange  consists  of  three  copies  of  the  same 
bill,  made  in  foreign  exchanges,  and  sent  by  different  conveyances 
to  provide  against  miscarriage ;  when  one  has  been  paid,  the  others 
are  void. 

592.  The  Face  of  a  bill  of  exchange  is  the  sum  ordered  to 
be  paid ;  it  is  usually  expressed  in  the  currency  of  the  place  on 
which  the  draft  is  made. 

093.  The  Par  of  Exchange  is  the  estimated  value  of  the 
coins  of  one  country  as  compared  with  those  of  another,  and  is 
either  intrinsic  or  commercial. 

094.  The  Intrinsic  Par  of  Exchange  is  the  comparative 
value  of  the  coins  of  different  countries,  as  determined  by  their 
weight  and  purity. 

090.  The  Commercial  Par  of  Exchange  is  the  comparative 
value  of  the  coins  of  different  countries,  as  determined  by  their 
nominal  or  market  price. 

NOTR. — The  intrinsic  par  is  always  the  same  while  the  coins  remain  un- 
changed; but  the  commercial  par,  being  determined  by  commercial  usage,  is 
fluctuating. 

096.  The  Course  of  Exchange  is  the  current  price  paid  in 
one  place  for  bills  of  exchange  on  another  place.  This  price 
varies,  according  to  the  relative  conditions  of  trade  and  commercial 
credit  at  the  two  places  between  which  exchange  is  made.  Thns, 
if  Boston  is  largely  indebted  to  Paris,  bills  of  exchange  on  Paris 
will  bear  a  high  price  in  Boston. 

When  the  course  of  exchange  between  two  places  is  unfavor- 


EXCHANGE.  3o9 

able  to  drawing  or  remitting,  the  disadvantage  is  sometimes 
avoided,  by  means  of 'a  circuitous  exchange  on  intermediate  places 
between  which  the  course  is  favorable. 

DIRECT   EXCHANGE. 

«597.  Direct  Exchange  is  confined  to  the  two  places  between 
which  the  money  is  to  be  remitted. 

598.  There  are  always  two  methods  of  transmitting  money 
between  two  places.     Thus,  if  A  is  to  receive  money  from  B, 

1st.  A  may  draw  on  B,  and  sell  the  draft; 
2d.  B  may  remit  a  draft,  made  in  favor  of  A. 

NOTE.  —  One  person  is  said  to  draw  on  another  person,  when  he  is  the  maker 
of  a  draft  addressed  to  that  person. 

CASE   I. 

599.  To  compute  domestic  exchange. 

The  course  of  exchange  ior  inland  bills,  or  drafts,  is  always  ex- 
pressed by  the  rate  of  premium  or  discount.  Drafts  on  time, 
however,  are  subject  to  bank  discount,  like  notes  of  hand,  for  the 
term  of  credit  given.  Hence;  their  cost  is  affected  by  both  the 
course  of  exchange  and  the  discount  for  time. 

1  What  will  be  the  cost  of  the  following  draft,  exchange  on 
Boston  being  in  Pittsburgh  at  2J  %"  premium? 

$600.  PITTSBURGH,  June  12,  1860. 

Sixty  days  after  sight,  pay  to  William  Barnard,  or  order,  six. 
hundred  dollars,  value  received,  and  charge  the  same  to  om 
account. 

To  the  Suffolk  Bank,  Boston.  THOMAS  BAUER  &  Co. 

OPERATION. 

$1  4-  $.0225  =  81.0225,  course  of  exchange. 

.0105,  bank  discount  of  $1,  (63  da.) 

$1.012,    cost  of  exchange  for  $1. 
X  1.012  =  $607.20,  Ans. 


AXALYSIS.  From  $1.0225,  the  course  of  exchange,  we  subtract 
$.0105,  the  bank  discount  of  $1  for  the  specified  time,  and  obtain 
£1.012,  the  cost  of  exchange  for  $1 ;  then  $600  X  1.012  =  $G07.20.  the 
cost  of  exchange  for  $600. 


340  PERCENTAGE. 

2.  A  commission  merchant  in  Detroit  wishes  to  remit  to  his 
employer  in  St.  Louis,  $512.86  by  draft  at  GO  days;  what  is  the 
face  of  the  draft  which  he  can  purchase  with  this,  sum,  exchange 

being  at  2J-  %  discount? 

OPERATION. 

$1  —  $.025  ==  $.975,       course  of  exchange. 
_.01225,  discount  of  $1. 

$.96275,  cost  of  exchange  for  $1. 
$512.36  -r-  .96275  =  $532.18  +  ,  Am. 

ANALYSIS.  From  $.975,  the  course  of  exchange,  we  mbtract 
$.01225,  the  bank  discount  of  $1  for  the  specified  time,  at  the  legal 
rate  in  Detroit,  and  obtain  $.96275,  the  cost  of  exchange  for  $1  ;  and 
the  face  of  the  draft  that  will  cost  $512.36,  will  be  as  many  dollars  as 
$.90275  is  contained  times  in  512.36,  which  is  532.18 -f,  times. 
Hence  we  have  the  following 

RULE.  I.  To  find  the  cost  of  a  draft,  the  face  being  given. — 
Multiply  the  face  of  the  draft  by  the  cost  of  exchange  for  $1. 

II.  To  find  the  face  of  a  draft,  the  cost  being  given. — Divide 
the  given  cost  by  the  cost  of  exchange  for  $1. 

NOTE. — The  cost  of  exchange  for  $1  may  always  be  found,  by  subtracting  from  the 
course  of  exchange  the  bank  discount  (at  the  legal  rate  where  the  draft  is  made), 
for  the  specified  time.  For  sight  drafts,  the  course  of  exchange  is  the  cost 
of  $1. 

EXAMPLES    FOR    PRACTICE. 

1.  What  must  be  paid  in  New  York  for  a  draft  on  Boston,  at 
30  days,  for  $5400,  exchange  being  ao  \  %  premium  ? 

Ans.  85392..%. 

2.  What  is  the  cost  of  sight  exchange  on   New  Orleans,  for 
$3000,  at  3}  %  discount? 

3.  What  must  be  paid  in  Philadelphia  for  a  draft  on  St.  Paul 
drawn    at   90    days,  for   $4800,  the  course   of  exchange   being 
1011  %?  Am.  $4791.60.     , 

4.  A  sight  draft  was  purchased  for  $550.62,  exchange  being  fit 
a  premium  of  31,  %  ;   \\liut  was  the  face  ? 

5.  An  agent  in  Syracuse,  N.  Y.,  having  $1324.74  due  his  em- 
ployer, is  instructed   to  remit  the  same  by  a  draft  drawn  at  30 
days;  what  will  be  the  face  of  the  draft,  exchange  being  at  If  % 
premium?  Ans.  $1310.22—. 


EXCHANGE.  341 

6.  My  agent  in   Charleston,   S.    C.,  sells  a  house  and  lot  for 
$7500,  on  commission  of  \\%,  and  remits  to  me  the  proceeds  in 
a  draft  purchased  at  \  %  premium ;  what  sum  do  I  receive  from 
the  sale  of  my  property  ? 

7.  A  man  in  Hartford,  Conn.,  has  $4800  due  him  in  Baltimore ; 
how  much  more  will  he  realize  by  making  a  draft  for  this  sum  on 
Baltimore  and  selling  it  at  \%  discount,  than  by  having  a  draft 
on  Hartford  remitted  to  him,  purchased  in  Baltimore  for  this  stun 
at  J  %  premium  ?  Arts.   $11.73  +  . 

8.  The  Merchants'  Bank  of  New  York  having  declared  a  dividend 
of  6J-  %,  a  stockholder  in  Cincinnati  drew  on  the  bank  for  the  sum 
due  him,  and  sold  the  draft  at  a  premium  of  1 J  (/c,  thus  realizing 
$508.75  from  his  dividend;  how  many  shares  did  he  own? 

9.  Sight  exchange  on  New  Orleans  for  $5000  cost  $5075  ;  what 
was  the  course  of  exchange?  Ans.   l£  fc  premium. 

CASE     II.* 

600.  To  compute  foreign  exchange. 

O01.  The  following  standards  of  the  decimal  currency  of  the 
United  States  were  established  by  the  coinage  act  of  1873 : 

Coins.  Weight.  Fineness. 

Gold  eagle 258       grains.  900  thousandths. 

Gold  dollar       ....  25.8        "  900  " 

Trade  dollar     ....  420  "  900  " 

Half-dollar 12^  grammes.  900          " 

Five-cent  (nickel).     .     .  77.16  grains.          f  copper,  £  nickel. 

Three-cent      "...  30  "  f-       u        i       " 

One-cent  (bronze) ...  48  "  .95       "  .05  tin  and  zinc. 

6O2.  Money  of  Account  consists  of  the  denominations  or 
divisions  of  money  of  any  particular  country  in  which  accounts  are 

kept. 

NOTE.— The  Act  of  March  3,  1P73,  provides  that  "the  value  of  foreign  coin,  as  ex- 
pressed in  thp  money  of  account  of  the  United  States,  uhall  be  that  of  the  pure  n:e!al  of 
such  coin  of  standard  value:  and  the  values  of  the  standard  coins  in  circnlifion.  of  the 
various  nation-*  of  the  world,  shnll  V  estimated  cm w ally  by  the  Director  of  the  Mint, 
and  be  proclaimed  on  the  first  day  of  January  by  the  Secretary  of  the  Treasury." 

*  This  "  CASE  "  on  foreirrn  exchange  has  been  so  modified  as  to  conform  to  the  Act 
of  March,  1873,  and  only  such  chances  have  been  made  in  the  Tables  and  Example.-  as 
were  necessary  to  adapt  them  to  that  lav/,  and  to  the  usage  of  1875. 


342 


PERCENTAGE. 


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EXCHANGE. 


343 


Weighty  Fineness,  'and  Value  of  Foreign  GOLD  Coins,   as 
determined  by  United  States  Mint  Assays. 


Country. 

Denomination. 

Weight. 

Fineness. 

Value  in  U.S. 
gold  coin. 

Austria 

Fourfold  ducat 

Ounces. 
0  448 

Thws'dtJis. 
986 

$  cfe.  m. 
9    13    2 

Do. 

Souverain  (no  longer  coined)  . 

0  363 

900 

6    75    4 

Do 

4  florins 

0  104 

900 

1    93    5 

Belgium  .             ... 

25  francs                      

0  254 

899 

4    72    0 

Brazil 

20  milreis 

0  575 

916  5 

10    89    4 

Ceuf'l  America 

2  escudos 

0  209 

853  5 

3    68    8 

Do.        do. 

4  reals  

0027 

875 

0    48    8 

Chili  
Colombia  and  S.  A. 
generally  .  . 

10  pesos  (dollars)  
Old  doubloon 

0.492 
0  867 

898 
870 

9    13    6 

15    50    3 

Denmark  

Old  10  thaler. 

0427 

895 

7    90    0 

Do  

New  20  crowns 

0  288 

900 

5    35    8 

Egypt  

Bedidlik  (100  piasters) 

0275 

875 

4    97    4 

England 

Pound  or  Sovereign  (new) 

02^8 

916  5 

4    86    5 

Do  
France 

Pound  average  (worn)  
20  franc  (no  new  issues) 

0,256,3 
0  207 

916,5 
899 

4    85    6 
3    84    7 

Germany 

Old  10  thaler  (Prussian) 

0'427 

903 

7    97    1 

Do  

New  20  marks  

0.256 

900 

4    76    2 

Greece 

20  drachms 

0  185 

900 

3    44    2 

India  (British)  .... 

Mohur  or  15  rupees         

0.375 

916.5 

7    10    5 

Italy 

20  lire  (francs) 

0207 

899 

3    84    7 

Japan 

Cobang  (obsolete)               .... 

0.289 

572 

3    57    6 

Do 

Xew  °0  yen 

1  072 

900 

19    94    4 

Mexico    

Old  doubloon  (average) 

0.867 

870 

15    59    3 

Do 

1  OP6 

875 

19    64    3 

Do  

20  pesos  (republic),  new  

1.081 

873 

19    51    5 

Netherlands  

10  guilders 

0.215 

899 

3    99    7 

New  Granada.. 

10  pesos  (dollars) 

0.525 

891,5 

9    67    5 

Peru  

20  soles        

1.055 

898 

19    21    3 

Portugal 

Coroa  (crown)               .... 

0308 

912 

5    80    7 

Russia 

0210 

916 

3    97    6 

Spain  . 

100  reales 

0268 

896 

4    96    4 

Do 

80  reales 

0215 

869,5 

3    86    4 

Do 

10  escudos 

0.270,8 

896 

5    01    5 

Sweden 

Ducat 

0111 

975 

2    23    7 

Do. 

Carolin  (10  francs)        

0.104 

900 

1    93    5 

Do 

New  20  crowns 

0288 

900 

5    35    8 

Tunis  

25  piasters  ....               .... 

0.161 

900 

2    99    5 

Turkey     .... 

100  piasters 

0231 

915 

4    37    0 

NOTES. — 1.  Foreign  gold  coins,  if  converted  into  United  States  coins, 
are  subject  to  a  charge  of  one-fifth  of  one  per  cent. 

2.  For  silver  coins  there  is  no  fixed  legal  valuation,  as  compared  with 
gold.  The  value  of  the  silver  coins  January  1, 1874,  was  computed  at  the 
rate  of  120  cents  per  ounce,  900  fine,  payable  in  subsidiary  silver  coin, 
or  113  cents  in  gold. 


344 


PERCENTAGE. 


STOCK     TABLE, 


Showing  the  rate  of  Interest  received  on  Stocks  purchased,  from  25  per 
cent,  discount  to  25  per  cent,  premium. 


Purchase 

BATE  RECEIVED  ON  STOCK  BEARING  INTEREST  AT 

Price. 

5  per  cent. 

6  per  cent. 

7  per  cent. 

8  per  cent. 

10  per  cent. 

75. 

6.666 

8.00 

9.333 

10.668 

13.333 

80. 

6.25 

7.50 

8.75 

10.000 

12.500 

85. 

5.882 

7.143 

8.235 

9.411 

11.764 

90. 

5.555 

6666 

7.777 

8.888 

11.111 

95. 

5.263 

6.316 

7.263 

8.421 

10.526 

97.50 

5.128 

6.156 

7.179 

8.205 

10.256 

100. 

5.000 

6.000 

7.000 

8.000 

10.000 

105. 

4.761 

5.714 

6.666 

7.619 

9.523 

110. 

4.545 

5454 

6.363 

7.272 

9.090 

115. 

4.347 

5.130 

6.086 

6.956 

8.695 

120. 

4.166 

5.000 

5.833 

6.660 

8.333 

125. 

4.000 

4.800 

5.600 

6.400 

8.000 

NOTES.— 1.  The  standard  value  of  gold  as  compared  with  silver  in  the  United  States, 
is  as  15.407  to  1  in  the  coinage  of  1792,  as  15.988  to  1  in  the  coinage  of  1837,  anrl  as  14.884 
to  1  in  the  coinage  of  1853.  By  the  new  Coinage  Act  of  1873,  there  is  no  fixed  legal  val- 
uation of  silver  as  compared  with  gold.  The  price  paid  at  the  mints  varies  according  to 
demand  and  supply. 

2.  The  relative  values  of  gold  and  silver  differ  in  the  coinage  of  different  countries. 
In  England,  the  ratio  is  14.288  to  1 ;  in  France,  it  is  15.5  to  1 ;  in  Prussia,  15.5  to  1. 

3.  Tn  the  gold  coinage  of  the  United  States,  a  Troy  ounce  of  fine  gold  is  equal  to 
$20.672 ;  and  an  ounce  of  standard  gold,  .900  fine,  at  the  legal  rate  of  25.8  grains  to  a  dol- 
lar, is  worth  $18.605. 


6O4.  Sterling:  Bills,  or  Sterling  Exchange,  are  bills  on  Eng- 
_and,  Ireland,  or  Scotland.  Such  bills  are  negotiated  at  a  rate  fixed 
without  reference  to  the  par  of  exchange. 

NOTE. — Formerly  such  hills  were  quoted  at  a  certain  rate  #  above  the  old  par  value  of 
a  pound  sterling,  which  was  $4.44  J.  As  this  was  entirely  a  fictitious  value,  and  always 
about  9  %  below  the  real  value,  the  course  of  exchange  always  appeared  1o  he  heavily 
against  this  country,  and  thus  tended  to  impair  its  credit,  By  the  Act  of  March,  1873, 
"all  contracts  made  after  the  first  day  of  January.  1874,  based  on  an  assumed  par  of  ex- 
change with  Great  Britain  of  fifty-four  pence  to  the  dollar,  or  $4.44 £  to  the  sovereign  or 
pound  sterling,"  are  declared  mill  and  roid.  The  pa.- of  exchange  between  Great  Britain 
and  the  United  States  is  fixed  at  $4.8665. 


EXCHANGE.  345 

6O5.  Exchanges  with  Europe  are  effected  chiefly  through  the 
following  prominent  financial  circles  :  London,  Paris,  Antwerp, 
Amsterdam,  Hamburg,  Frankfort,  Bremen,  and  Berlin. 

NOTE.—  1.  In  exchange  on  Paris,  Antwerp,  and  Switzerland,  the  unit  is  the  franc,  and 
the  quotation  shows  the  number  of  francs  and  centimes  to  the  dollar,  Federal  Money. 
In  exchange  on  Amsterdam,  the  unit  is  the  guilder,  quoted  at  its  \alue  in  cents;  on 
Hamburg,  Frankfort,  Bremen,  and  Berlin,  the  quotation  shows  the  value  of  four  reichs- 
marks  (marks)  iii  cents. 

2.  The  following  shows  the  manner  in  which  quotations  of  foreign  exchange  are 
made  in  this  country,  and  as  quoted  Jan.  2,  1875: 


London  
Do 

.  .Prime  Bankers'  Sterling  Bills.  .  . 

Sixty  Days. 
..   4.85%  @  4.86 

Three  Days. 
4.90     @  4.!tO'^ 

Do  
Paris  

.  .  .  Prime  Commercial       do. 
.  Francs.  ... 

..   4.84     ©4.85 
5  17>£  (a;  5  16>£ 

4.881/2  ©4.89^ 
5  13*^  (ff,  5  12  JX 

Antwerp   .  .  . 

...Francs  ..          

.  .    5.17J4  ®  5  IG1^ 

5  13^  tjy  5  121/;; 

Switzerland. 

.  .  Francs 

5  n%  @  5  ig^- 

5  13V  ©  5  12i/ 

Amsterdam 

Guilders 

41  Vg  @.     41  14 

4ij^  @,    41  5' 

Hambur^ 

Reichsmarks       .       ..        .... 

94%  ©     95)s 

96     ©    %j^ 

Frankfort 

947^  (g,     951^ 

c.g     (^     96^ 

Bremen  .  .  . 

Reichsmarks            

94  7s  @     95% 

96     ©     %X 

Berlin... 

...Reichsmarks  ... 

9472  (fh     95^ 

96     Gh     96^ 

In  the  above,  u  Prime  Bankers1  Bills"  are  those  on  the  most  reliable  banking  houses; 
"Good  'Ms  applied  to  those  of  somewhat  inferior  credit.  :  and  u  Prime  Commercial" 
are  merchants'  drafts,  which  usually  command  a  less  price  in  the  market.  The  quota- 
tions in  the  first  column  are  those  of  60-day  bills,  and  in  the  second  column  those  of 
3  days. 

1.  What  will  be  the  cost  in  Boston  of  the  following  bill  of  ex- 
change on  Liverpool,  the  course  of  exchange  being  4.8  7-g-  ? 

£432.  BOSTON,  June  16,  1875. 

At  sight  of  this  First  of  Exchange  (Second  and  Third  of  same 
tenor  and  date  unpaid),  pay  to  the  order  of  J.  Simmons,  Boston, 
Four  Hundred  Thirty-two  Pounds,  value  received,  and  charge  the 
same  to  account  of 

JAMES  LOWELL  &  Co. 
To  RICHARD  EVANS  &  SON,  ) 
Liverpool,  England.          f 

OPERATION.  ANALYSIS.—  According 

$4.875,  course  of  exchange,  value  of  £1.      to    the    course     of    ex- 

432,  number  of  pounds.  change,     £1     is    worth 

$4.875  x  432  =$2106,  Ans.  $4.875;    hence    £432  is 

worth  432  times  $4.875, 
or  $2106,  the  required  cost  of  the  bill. 


346  PERCENTAGE. 

2.  What  is  the  face  of  a  bill  on  London,  that  may  be  purchased 
in  New  York  for  $277.42,  exchange  being  quoted  at  4.85? 

OPERATION.  ANALYSIS. — Since  £1  costs  $4.85,  as 

$977  42  —  $4  85  =  57  2        many    pounds    can     be     bought    for 

$277.42  as  $4.85  is  contained  times  in 
£57.2  =  £57  4s.,  the  face.      $277.42,  or £57.2=£57  4s.    Hence,  etc. 

3.  What  cost  in  Hamburg  a  bill  on  New  Orleans  for  $4500,  the 
course  of  exchange  being  95  ? 

OPERATION.  ANALYSIS. — Since  $.95  is 

$4500  -T-  $.95  =  4736.84.  tlle  cost  of  4  marks  (6O5), 

4736.84  x  4  =  18947.36  marks.          $4500  wil1  cost  four  times 

18947  marks  36  pennies,  cost  of  bill.      as  man?  marks  as  $-95  is 

contained  times  in  $4500,  or 

18947.36.     Hence  the  cost  of  the  bill  is  18947  marks  36  pennies. 

GOO.  From  these  illustrations  we  derive  the  following 

RULE.  I.  To  find  the  cost  of  a  bill,  the  face  being  given. — 
Multiply  the  value  of  a  monetary  unit  according  to  the  course  of 
exchange,  by  the  number  of  such  units  in  the  face  of  the  bill. 

II.  To  find  the  face  of  a  bill,  the  cost  being  given. — Divide  the 
cost  of  the  bill  by  the  value  of  each  monetary  unit,  according  to  the 
course  of  exchange. 

EXAMPLES     FOR     PRACTICE. 

1.  What  is  the  cost  in  Portland  of  a  bill  on  Manchester,  Eng., 
for  £325  3s.  9d.,  when  sterling  exchange  is  selling  at  4.89£? 

Ans.  $1591.79  +  . 

2.  What  must  be  paid  in  Charleston  for  a  bill  of  exchange  on 
Paris  for  6000  francs,  the  course  of  exchange  being  5.31  ? 

NOTE.— The  quotation  5.31  means  that  number  of  francs  to  a  dollar.  Hence  f  1.00-s- 
5.31  gives  the  value  of  1  franc,  which  multiplied  by  the  number  of  francs  (by  Rule  I)  will 
give  the  cost  of  the  bill ;  or  more  briefly,  the  given  number  of  francs  divided  by  the 
quotation  will  give  the  cost  of  the  bill  in  dollars. 

3.  What  is  the  cost,  m  Boston,  of  a  bill  on  St.  Petersburg  for 
3000  roubles,  at  $.771;  brokerage  \%r\     Find  also   at  what  per 
?cnt.  premium  the  exchange  is  at  that  time.      (See  Table,  p.  342.) 

4.  What  will  be  the  cost  in  Naples  of  a  bill  of  exchange  on  New 
York  for  $831.12,  at  $.205  a  lira? 


EXCHANGE.  347 


5.  A  draft  on  Philadelphia  cost  £125  in  Birmingham, 
exchange  selling  for  4.855;  required  the  face  of  the  draft. 

Ans.  $606.875. 

6.  An  agent  in  Boston  having  $7536.30  due  his  employer  in 
England,  is  directed  to  remit  by  a  bill  on  Liverpool  ;  what  is  the 
face  of  the  bill  which  he  can  purchase  for  this  money,  exchange 
selling  at  4.91  £?  Ans.  £1534  2s.    Id.  + 

7.  A  merchant  in  Cincinnati  has  9087-J  guilders  due  him  ,in 
Amsterdam,  and  requests  the  remittance  by  draft.     What  sum  will 
he  receive,   exchange  on   U.    S.    in    Amsterdam  selling  at  2.41 
guilders  for  $1  ? 

8.  A  trader  in  London  wishes  to  invest  £2500  in  merchandise  in 
Lisbon  ;  if  he  remits  to  his  correspondent  at  Lisbon  a  bill  purchased 
for  this  sum,  at  the  rate  of  4.51  milreis  to  the  pound  sterling,  what 
sum  in  the  currency  of  Portugal  will  the  agent  receive  ? 

Ans.  11275  milreis. 

9.  A  draft  on  Dublin  for  £360  cost  $1736.10;  what  was  the 
course  of  exchange  ?  Ans.   4.82  J. 

10.  A  merchant  in  Baltimore  having  received  an  importation  of 
Madeira  wine  invoiced  at  1  500  milreis,  allows  his  correspondent  in 
Madeira  to  draw  on  him  for  the  sum  necessary  to  cover  the  cost, 
exchange  on  the  United  States  being  in  Madeira  931  £  reis  to  the 
dollar;  how  much  would  the  merchant  have  saved  by  remitting  a 
draft  on  Madeira,  purchased  at  $1.065  per  milreis? 

Ans.  $12.80  +  . 

11.  An  importer  received  a  quantity  of  Leghorn  hats,  invoiced 
at  25256.80  lire,  which  was  paid  in  U.  S.  gold  coin,  exported  at  a 
cost  of  3  %  for  transportation  and  insurance,  the  price  of  fine  gold 
in  Leghorn  being    131    lire    per  ounce  Troy.     How  much  more 
would  the  goods  have  cost  in  store,  had  payment  been  made  by 
draft  on  Leghorn,  purchased  at  the  rate  of  $.198  per  lira? 

Ans.  $895.75  -f  . 
NOTE.—  In  U.  S.  gold  coinage,  $10  contains  258  x  .9=232.2  grains  offlne  gold  (603). 

12.  When  silver  is  worth  in  England  61d.  per  oz.  fine,  what  sum 
of  money  in  U.  S.  trade-dollars  is  equal  to  £l  sterling  ? 

Ans.   $4.996  +  . 

13.  WTiat  is  the  course  of  exchange  on  Berlin  when  $858.85  ia 
paid  for  890  marks? 


PERCENTAGE. 


6O7.  Arbitration  of  Exchange  is  the  process  of  computing 
exchange  between  two  places  by  means  of  one  or  more  intermediate 
exchanges.  < 

NOTES.— When  there  is  only  one  intermediate  exchange,  the  process  is  called  Simple 
Arbitration  ;  when  there  are  two  or  more  intermediate  exchangee,  the  process  is  called 
Compound  Arbitration 

2.  The  arbitrated  price  is  generally  either  greater  or  less  than  the  price  of  direct  ex- 
changes ;  and  the  object  of  arbitration  is  to  ascertain  the  best  route  for  making  draita 
or  remittances. 

,GO8.  There  are  always  three  methods  of  receiving  money  from 
a  place,  or  of  transmitting  money  to  a  place,  by  means  of  indirect 
exchange  through  one  intervening  place.  Thus, 

If  A  is  to  receive  money  from  C  through  B, 

1st.  A  may  draw  on  B,  and  B  draw  on  C; 

2d.  A  may  draw  on  B,  and  C  remit  to  B ; 

3d.   B  may  draw  on  C,  and  remit  to  A. 

If  A  is  to  transmit  money  to  C  through  B, 

1st.  A  may  remit  to  B,  and  B  remit  to  C ; 

2d.  A  may  remit  to  B,  and  C  draw  on  B ; 

3d.  B  may  draw  on  A,  and  remit  to  C. 

1.  A  man  in  Albany,  N.  Y.,  paid  a  demand  in  Paris  of  5400  fr., 
by  remitting  to  Amsterdam  at  the  rate  of  1.41-J  per  guilder,  and 
thence  to  Paris  at  the  rate  of  2.15  francs  per  guilder.  How  much 
Federal  money  was  required  ? 


OPEKATION. 

$  (?)  —  5400  francs. 

2.15  francs    =        1  guilder. 
1  guilder  =  .8.41  J. 

(?)  =  $1042.32+,  Ans. 
Or, 


ANALYSIS.— We  are  to  deter- 
mine how  much  Federal  money 
is  equal  to  5400  francs,  and  the 
question  may  be  represented 
thus:  $(?)=5400francs.  Now, 
since  2.15  francs  =  1  guilder, 
5400  divided  by  2.15  will  give  the 
number  of  guilders;  and  that 
number  multiplied  by  $.415,  the 
value  of  1  guilder,  will  give  the 
required  sum.  Hence,  5400  and 
.415  are  multipliers  and  2.15  is  a 
divisor.  The  units  of  currency 
being  canceled,  and  the  work 

being;  abridged  also  by  canceling  common  factors,  we  have     (?)  = 

$1042.32  +  ,  the  sum  required. 


(?) 

5400 

.43  JU$ 

1 

1 

M$  .083 

.43 

$448.2 

$1042.32  +  ,  Ans. 

EXCHANGE.  349 

Or,  since  the  course  of  exchange  between  Amsterdam  and  Paris  gives 
1  guilder=2.15  francs^  and  the  course  between  Albany  and  Amsterdam 
gives  $.41^=1  guilder,  we  multiply  the  5400  francs  by  .415  and  divide 
by  2.15,  using  the  vertical  line  and  cancellation,  and  obtain  $1042.32  +  , 
as  before. 

NOTE.—  In  the  first  statement,  the  rates  of  exchange  are  so  arranged  that  the  same 
unit  of  currency  shall  stand  on  opposite  sides  in  each  two  consecutive  equations,  in 
order  that  these  factors  may  all  be  canceled. 

2.  A  resident  of  Naples,  having  a  bequest  of  $8720  made  him  in 
Boston,  orders  the  remittance  to  be  made  to  his  agent  in  London, 
who  remits  the  proceeds  to  Naples,  reserving  his  commission  of  -J-^ 
on  the  draft  sent.  If  the  course  of  exchange  on  London  is  8-J-.875 
in  Boston,  and  the  rate  between  London  and  Naples  is  25.53  lire  to 
the  pound  sterling,  how  much  does  the  man  realize  from  his  bequest  ? 

OPERATION.  ANALYSIS.  —  We  make  the  statement  as 

(?)  lire  =  $8720  m  ^le  ^rst   examP^e>  according  to  the 

<$;4  07^    -  -  £i  given  rates  of  exchange.     Then,  since  the 

'""  *  agent  is  to  deduct  $  %  commission  on  the 

1,5.53    ire. 


of  the  draft  befofe  the     urcnase    we 

05     —  1 


place  1.005  on  the  left  as  a  divisor  (152), 
(?)  =  45438.77  lire,  and  obtain  by  cancellation,  multiplica- 

tion, and  division,  45438  lire  77  centimes 
as  the  proceeds  of  the  exchange. 

3.  A  merchant  in  Chicago  directs  his  agent  in  Albany  to  draw 
upon  Baltimore  at  1  %  discount,  for  81200  due  from  the  sales  of 
produce;  he  then  draws  upon  the  Albany  agent,  at  2  %  premium, 
for  the  proceeds,  after  allowing  the  agent  to  reserve  -J  %  for  his 
commission.  What  sum  does  the  merchant  realize  from  his  produce  ? 

OPERATION.  ANALYSIS.  —  According    to   the    given 

/  M  Q  __  i9oo  B  rates  of  exchange,  100  dollars  in  Balti- 

j  QQ  g  _       QQ  A  more  equal   99  dollars  in  Albany  ;   and 

100  A  -      10^  r*  *^  dollars  in  Albany  equal  102  dollars 

in  Chicago  ;  and  since  the  unit  of  currency 

=  is  the  same  in  each  place,  being  $1,  we 


(  ?  )  =  $1205.70,  Ans.  represent  its  exchange  value  in  each  town 
by  the  initial  letter,  and  make  the  state- 
ment as  in  the  other  examples.  Then,  since  the  agent  is  to  reserve  \  % 
commission  from  the  avails  of  his  draft,  we  place  1— . 005  =  .  995  on  the 
right  as  a  multiplier,  and  obtain  by  cancellation  (?)  =  $1205.70,  the 
answer. 


350  PEKCENTACE. 

From  these  principles  and  illustrations  we  have  the  following 

RULE.  I.  Represent  the  required  sum  by  (  ?  )f  with  the  proper 
unit  of  currency  affixed,  and  place  it  equal  to  the  given  sum  on  the 
right. 

II.  Arrange  the  given  rates  of  exchange  so  that  in  any  two  con- 
secutive equations  the  same  unit  of  currency  shall  stand  on  opposite 
sides. 

III.  When  there  is  commission  for  drawing,  place  1  minus  the 
rate  on  the  left  if  the  cost  of  exchange  is  required,  and  on  the  right 
if  proceeds  are  required  ;  and  when  there  is  commission  for  remit- 
ting, place  1  plus  the  rate  on  the  right  if  cost  is  required,  and  on  the 
left  if  proceeds  are  required. 

IV.  Divide  the  product  of  the  numbers  on  the  right  by  the  product 
of  the  numbers  on  the  left,  cancelling  equal  factors  ;  the  result  will 
be  the  answer. 

NOTES.— 1.  Commission  for  drawing  is  commission  on  the  sale  of  a  draft;  commis- 
sion for  remitting  is  commission  on  the  purchase  price  of  a  draft. 

2.  The  above  method  is  sometimes  called  the  Chain  Rule,  or  Conjoined  Proportion. 

EXAMPLES     FOR     PRACTICE. 

1.  A  gentleman  in  Philadelphia  wishes  to  deposit  $5000  in  a  bank 
at  Stockholm,  by  remitting  to  Liverpool  and  thence  to  Stockholm ; 
if  the  course  of  exchange  on  Liverpool  is  4.91  in  Philadelphia,  and 
the  course  between  Liverpool  and  Stockholm  is  18£  crowns  to  £l, 
how  much  money  will  the  man  have  in  bank  at  Stockholm,  allow- 
ing the  agent  at  Liverpool  J^  for  remitting. 

Ans.  18792.1  crowns. 

2.  When  exchange  at  New  York  on  Paris  is  5  francs  16  centimes 
per  $1,  and  at  Paris  on  Hamburg  1.23  fr.  per  mark,  what  will  be 
the  arbitrated  price  in  New  York  of  7680  marks  of  Hamburg? 

Ans.  $1830.69. 

3.  A  gentleman  in  Cleveland  wishes  to  draw  on  New  Orleans  for  a 
bank  stock  dividend  of  $750,  and  exchange  direct  on  New  Orleans 
is  \\%  discount;  how  much  will  he  save  by  drawing  on  his  agent 
in  New  York  at  1  -J-  %  premium,  allowing  his  agent  to  draw  on  New 
Orleans  at  1  %  discount,  brokerage  at  £  %  ? 

4.  A  gentleman  in  Boston  drew  on  Amsterdam  for  6000  guilders 
at  $.415  per  guilder;  how  much  more  would  he  have  received  if  he 


EXCHANGE.  351 

had  ordered  remittance  to  London,  and  thence  to  New  York,  ex- 
change at  Amsterdam  on  London  being  11.19  guilders  per  £l,  and 
at  London  on  New  York  4.88,  brokerage  at  lj$  in  London  for 
remitting.  Ans.  $94.31+. 

5.  If  at  Philadelphia  exchange  on  Liverpool  is  4.89J-,  and  at 
Liverpool  on  Paris  24  francs  96  J  centimes  per  £l,  what  is  the  arbi- 
trated course  of  exchange  between  Philadelphia  and  Paris,  through 
Liverpool?  Ans.  5.10. 

6.  An  American  resident  of  Amsterdam  wishing  to  obtain  funds 
from  the  U.  S.  to  the  amount  of  $6400,  directs  his  agent  in  Lon- 
don to  draw  on  the  U.  S.  and  remit  the  proceeds  to  him  in  a  draft 
on  Amsterdam,  exchange  on  the  U.  S.  being  at  4.85  in  London, 
and  the  course  between  London  and  Amsterdam  being  18d.  per 
guilder.     If  the  agent  charges  commission  at  -£•%  both  for  drawing 
and  remitting,  how  much  better  is  this  arbitration  than  to  draw 
directly  on  the  U.  S.  at  41  cents  per  guilder? 

7.  A  speculator  in  Pittsburgh,  having  purchased  58  shares  of 
railroad  stock  in  New  Orleans,  at  95  %,  remits  to  his  agent  in  New 
York  a  draft  purchased  at  2  %  premium,  with  orders  for  the  agent 
to  remit  the  sum  due  in  N.  O.     Now,  if  exchange  on  N.  O.  is  at 
%fo  discount  in  N.  Y.,  and  the  agent's  commission  for  remitting  is 
J  (.f.  how  much  does  the  stock  cost  in  Pittsburgh  ?   Ans.  $5606.08. 

8.  A  merchant  in  Boston  owes  19570  francs  in  Paris.     Which 
will  be  the  more  advantageous  to  him,  to  remit  directly  to  Paris  at 
5.12  or  through  London  at  4.89,  buying  there  exchange  on  Paris  at 
25.19  fr.  to  £1,  and  paying  J^  brokerage? 

9.  If  in  London  exchange  on  Paris  is  25.71,  and  in  New  York  on 
Paris  it  is  5.1 5  J,  what  is  the  arbitrated  course  of  exchange  between 
New  York  and  London ?  Ans.  4.987+. 

10.  A  banker  in  New  York  remits  $3000  to  Liverpool,  by  arbitra- 
tion, as  follows:  first  to  Paris  at  5  francs  16  centimes  per  $1 ; 
thence  to  Hamburg  at  125  francs  per  100  marks;  thence  to  Am- 
sterdam at  1.71-J  marks  to  the  guilder;  thence  to  Liverpool  at  11.82 
guilders  per  £1  sterling.     How  much  sterling  money  will  he  have 
in  bank  at  Liverpool,  and  what  will  be  his  gain  over  direct  exchange 
at  4.91  £?  ^ns     j  Proceeds  in  Liverpool,  £610  18s.  3d. 

(  Gain  by  arbitration,      10s.  9d. 


352  PERCENTAGE. 

EQUATION  OF  PAYMENTS. 

609.  Equation  of  Payments  is  the  process '  of  finding  the 
mean  or  equitable  time  of  payment  of  several  sums,  due  at  dif- 
ferent times  without  interest. 

610.  The  Term  of  Credit  is  the  time  to  elapse  before  a  debt 
becomes  due. 

611.  The  Average  Term  of  Credit  is  the  time  to  elapse  before 
several  debts,  due  at  different  times,  may  all  be  paid  at  once,  with- 
out loss  to  debtor  or  creditor. 

613.  The  Equated  Time  is  the  date  at  which  the  several 
debts  may  be  canceled  by  one  payment. 

613.  To  Average  an  Account  is  to  find  the  mean  or  equit- 
able time  of  payment  of  the  balance. 

614.  A  Focal  Date  is  a  date  with  which  all  the  others  are  com- 
pared in  averaging  an  account. 

NOTE.  —  Each  item  of  a  book  account  draws  interest  from  the  time  it  is  due, 
which  may  be  either  at  the  date  of  the  transaction,  or  after  a  specified  terui  of 
credit. 

In  averaging,  there  are  two  kinds  of  equations,  Simple  and 
Compound. 

615.  A  Simple  Equation  is  the  process  of  finding  the  aver- 
age time  when  the  payments  or  account  contains  only   one  side, 
which  may  be  either  a  debit  or  credit. 

616.  A  Compound  Equation  is  the  process   of  averaging 
when  both  debts  and  credits  are  to  be  considered. 

SIMPLE   EQUATIONS. 
CASE    I. 

617.  When  all  the  terms  of  credit  begin  at  the  same 
date. 

1.  In  settling  with  a  creditor  on  the  first  day  of  April,  I  find 
that  I  owe  him  $12  due  in  5  months,  $15  due  in  2  months,  and 
$18  due  in  10  months ;  at  what  time  may  I  pay  the  whole  amount? 


EQUATION  OF  PAYMENTS.  353 

OPERATION.  ANALYSIS.     The 

$12  X     5  —     GO  "vrhole   amount   to    be 

15  X     2=     30  paid,  as  seen  in  the  ope- 

lp  x  10  =  180  ration,  is  $45  ;  and  we 

oj^  .^YQ  are  to  find  how  long  it 

•270 ---45  ="6  mo.,  average  credit,         sha11  be  withheld,  or 

Apr.  1,  +  6  mo.  =  Oc.t.  1,  Ans.  what  term  of  credit  It 

shall  have,  as  an  equiv- 
alent for  the  various  terms  of  credit  on  the  different  items.  Now 
the  value  of  credit  on  any  sum  is  measured  by  the  product  of  the 
money  and  time.  Therefore,  the  credit  on  $12  for  5  mo.  =  the  credit 
on  $00  for  1  mo.,  because  12  X  5  =  60  X  1.  In  like  manner,  we  have 
the  credit  on  $15  for  2  mo.  =  the  credit  on  $30  for  1  mo. ;  and  the 
credit  on  $18  for  10  mo.  =  the  credit  on  $180  for  1  mo.  Hence,  by 
addition,  the  value  of  the  several  terms  of  credit  on  their  respective 
sums  equals  a  credit  of  1  month  on  $270  ;  and  this  equals  a  credit  of 
6  months  on  $45,  because  45  X  6  =  270  X  1.  Hence  the  following 

E,ULE.  1.  Multiply  each  payment  l>y  its  term  of  credit,  and 
divide  the  sum  of  the  products  by  the  sum  of  the  payments'  the 
quotient  will  be  the  average  term  of  credit. 

II.  Add  the  average  term  of  credit  to  the  date  at  which  all 
the  credits  begin;  the  result  will  be  the  equated  time  of  payment. 

NOTES.  —  1.  The  periods  of  time  used  as  multipliers  must  nil  be  of  the  same 
denomination,  and  the  quotient  will  be  of  the  same  denomination  as  the  terms 
of  credit;  if  these  be  months,  and  there  be  a  remainder  after  the  division,  con- 
tinue the  division  to  days  by  reduction,  always  taking  the  nearest  unit  in  the  last 
result. 

2  The  several  rules  in  equation  of  payments  are  based  upon  the  principle  of 
bank  discount;  for  they  imply  that  the  discount  of  a  sum  paid  before  it  is  due 
equals,  the  interest  of  the  same  amount  paid  after  it  is  due. 

EXAMPLES    FOR    PRACTICE. 

1.  On  the  first  day  of  January,  1860,  a  man  gave  3  notes,  the 
first  for  $500  payable  in  30  days;  the  second  for  $400  payable  in 
60  days;  the  third  for  $600  payable  in  90  days.      What  was  the 
average  term  of  credit,  and  what  the  equated  time  of  payment? 

An  A    Term  of  credit,  62  da.;  time  of  payment,  Mar.  3,  1860. 

2.  A  man  purchased  real  estate,  and  agreed  to  pay  J  of  the  price 
in  3  mo.,  t  in  8  mo.,  and  the  remainder  in  1  year.      Wishing  to 
cancel  the  whole  obligation  at  a  single  payment,  how  long  shall 
this  payment  be  deferred  ? 


364 


PERCENTAGE. 


3.  I  owe  $480  payable  in  90  days,  and  $320  payable  in  60  days. 
My  creditor  consents  to  an  extension  of  time  to  1  year,  and  offers 
to  take  my  note  for  the  whole   amount  on   interest  at  6  per  cent, 
from  the  equated  time,  or  a  note  for  the  true  present  worth  of 
both  debts,  on  interest  from  date-     How  much  will  I  gain  if  I 
choose  the  latter  condition  ?  Ans.  $1.14. 

4.  Bought  merchandise  .April  1,  as  follows:  $280  on  3  mo., 
$300  on  4  mo.,  $200   on  5  mo.,  $560   on  6  mo. ;  what  is  the 
equated  time  of  payment  ?  Ans.  Aug.  24. 

CASE    II. 

618.  When  the  terms  of  credit  begin  at  different 
dates. 

1.  When  does  the  amount  of  the  following  bill  become  due, 
per  average  ? 

CHARLES  CROSBY, 

1860.  To  BRONSON  &  Co.,  Dr. 

Jan.  12.     To  Mdse., $400 

"     16.      "    Mdse.  on  2  mo., 600 

Apr.20,      "   Cash, ,  375 


PIRST    OPERATION. 


SECOND    OPERATION. 


Due 

Da. 

Items. 

Prod. 

Jan.    12 
Mar.  16 
Apr.  20 

64 
99 

400 
600 
375 

38400 
37125 

1375 

75525 

Due. 

Da. 

Items. 

Prod. 

Jan.  12 
Mar.  16 
Apr.  20 

99 
35 
0 

400 

600 
375 

39600 
21000 

1375 

.60600 

75525  -:-  1375  =  55  da. 
55  da.  after  Jan.  12, 
or  Mar.  7. 


60600  H-  1375  =  44  da. 
44  da,  before  Apr.  20, 
or  Mar.  7. 


ANALYSIS.  The  three  items  of  the  bill  are  due  Jan.  12,  Mar.  16, 
and  Apr.  20,  respectively.  In  the  first  operation  we  use  the  ear/ text 
maturity,  Jan.  12,  for  a  focal  date,  and  find  the  difference  in  days 
between  this  date  and  each  of  the  others ;  thus,  from  Jan.  12  to  Mar. 


EQUATION  OF  PAYMENTS.  355 

16  is  64  da. ;  from  Jan.  12  to  Apr.  20  is  99  da.  Hence,  from  Jan.  12 
the  first  item  has  no"  credit,  the  second  nas  64  days  credit,  and  the 
third  99  days'  credit,  as  appears  in  the  column  marked  da.  We  now 
proceed  to  find  the  products  as  in  Case  I,  whence  we  obtain  the  ave- 
rage credit,  55  da.,  and  the  equated  time,  Mar.  7. 

In  the  second  operation,  the  latest  maturity.  Apr.  20,  is  taken  for  a 
focal  date,  and  the  work  may  be  explained  thus :  Suppose  the  account 
to  be  settled  Apr.  20.  At  that  time  the  first  item  has  been  due  99 
days,  and  must  therefore  draw  interest  for  this  time.  But  interest 
on  $400  for  99  days  =  the  interest  on  $39600  for  1  day.  The  second 
item  must  draw  interest  35  days  ;  but  interest  on  $600  foi  35  days  = 
interest  on  $21000  for  1  day.  Taking  the  sum  of  the  products,  we  find 
that  the  whole  amount  of  interest  due  Apr.  20  equals  the  interest  on 
$60600  for  1  day ;  and  this  is  found,  by  division,  equal  to  the  interest 
on  $1375  for  44  da.,  which  is  the  average  term  of  interest.  Hence 
the  account  would  be  settled  Apr  20,  by  paying  $1375,  with  interest 
on  the  same  for  44  days.  .This  shows  that  the  $1375  has  been  used 
44  days,  that  is,  it  falls  due  Mar  7,  without  interest.  Hence  we  have 
the  following 

RULE.  I.  Find  the.  time  at  which  each  item  becomes  due,  by 
adding  to  the  date  of  each  transaction  the  term  of  credit,  if  any  be 
specified,  and  write  these  dates  in  a  column 

II.  Assume  either  the  earliest  or  the  latest  date  for  a  focal  date, 
and  find  the  difference  in  days  between  the  focal  date  and  each  of 
the  other  dates,  and  write  the  results  in  a  second  column. 

III.  Write  the  items  of  the  account  in  a  third  column,  and  mul- 
tiply each  by  the  corresponding  number  of  days  in  the  preceding 
column,  writing  the  products  in  a  fourth  column. 

IV.  Divide  the  sum  of  the  products  by  the  sum  of  the  items. 
The  quotient  will  be  the  average  term  of  credit  or  interest,  and 
must  be  reckoned  from  the  focal  date  TOWARD  the  other  dates,  to 
find  the  equated  time  of  payment. 

NOTES.— 1.  When  dollars  and  certs  are  given,  it  is  generally  sufficient  to  take 
only  dollars  in  the  multiplicand,  rejecting  the  cents  when  less  than  50,  and  carrying 
1  to  the  dollar?,  if  the  cents  are  more  than  50. 

2.  Months  in  any  terms  of  credit  are  understood  to  he  calendar  months ;  the  time 
must  therefore  he  carried  forward  to  the  same  day  of  the  month  in  which  the  term  of 
credit  expires. 


356  PERCENTAGE. 

EXAMPLES  FOR  PRACTICE. 

1.  JAMES  GORDON, 

1860.                                        To  HENRY  LANCEY,  Dr. 
Mar.    4.     To    100  yd.  Cassimere,      @  $2  50, $250 

"      25.      "   300C  <<•'    French  Prints,"       .12 360 

Apr.  16.      «   1200    "    Sheeting,         "       .08, 96 

"     30.      «     400    «    OilCloth,        «       .50, 200 

May  17.      "     Sundries, 350 

When  is  the  above  bill  due,  per  average  ? 

Am.  Apr.  12,  1860. 

2.  I  sell  goods  to  A  at  different  times,  arid  for  different  terms 
of  credit,  as  follows : 

Sept.  12,  1859,  a  bill  on  30  days'  credit,  for  $180 

Oct.      7,  "         "  30  "  «  300 

Nov.   16,  "         "  60  "  «  150 

Dec.   20,  «  «  90  "  «               350 

Jan.    25,  1860,  «  30  «  "               130 

Feb.   24,  "  "  30  "  "               140 
If  I  take  his  note  in  settlement,  at  what  time  shall  interest 
commence  ? 

3.  What  is  the  average  of  the  following  account  ? 

1860,  Oct.    1.  Mdse.,  on  60  da............ $240 

"       Nov.  12.        "         «       "        500 

"       Dec.  25.        "         "       "       436 

1861,  Jan.  16.        "         "       <"      325 

"       Feb.  24.        "         «       "       436 

«       Mar.  17.        «         «       "        537 

Atis.   Mar.  10,  1861. 

4.  I  have  4  notes,  as  follows:  the  first  for  $350,  due  Aug.  16, 
1859  •  the  second  for  8250,  due  Oct.  15, 1859 ;  the  third  for  3r)>00, 
due  Dec.    14,   1859;  the  fourth  for  $248,  due  Feb.  12,   1860. 
When  shall  a  note  for  which  I  may  exchange  the  four,  be  made 
payable  ? 


EQUATION  OF  PAYMENTS. 


35" 


Dr. 


COMPOUND    EQUATIONS. 

O1O.    1    Average  the  following  account. 
John  Lyman. 


Or. 


18fiO. 

1360. 

June  12 

To  Mdse. 

530 

00  |    June  24  |     Bv  draft  at  30  da. 

480 

00 

Sept.   12 

U                 fi 

428 

00  1    Aug.   20  j      ••    cash, 

280 

00 

Oct.      28 

"   Sundries, 

440 

00  II  Oct.       81      «       •' 

140 

00 

OPERATION. 


Dr. 


Cr. 


Focal 
date, 


Due. 

June  12 
Sept.   12 
Oct.     28 

Da. 

Items. 

Products. 

Due. 

Da. 

Items. 

Products. 

138 
46 
0 

530 
428 
440 

73140 
19688 

July  27 
Aug.  20 
Oct.     8 

93 
69 
20 

480 
230 
140 

44640 
15S70 
2800 

Balances, 

1398 
850 

548 

9-2828 
63310 

850 

63310 

29518 

29518  -f-  548  =  54  da.,  average  term  of  interest. 
Oct.  28  —  54  da.  =  Sept.  4,  balance  due. 

ANALYSIS. — In  this  operation  we  have  written  the  dates  of  maturity 
on  either  side,  allowing  3  days'  grace  to  the  draft.  The  latest  date, 
Oct.  28,  is  assumed  as  the  focal  date  for  both  sides,  and  the  two  columns 
marked  da.  show  the  difference  in  days  between  the  focal  date  and 
each  of  the  other  dates.  The  products  are  obtained  as  in  simple 
equations,  and  the  balance  found  between  the  items  on  the  two  sides, 
and  also  between  the  products.  These  balances,  being  both  on  the 
Dr.  side,  show  that  there  is  due  on  the  day  of  the  focal  date,  $548, 
with  interest  on  $29518  for  1  day.  By  division,  this  interest  is  found 
to  be  equal  to  the  interest  on  $548  for  54  days.  Hence  this  balance, 
$548,  has  been  due  54  days ;  and  reckoning  back  from  the  focal  date, 
we  obtain  the  equated  time  of  payment,  Sept.  4. 

Had  we  taken  the  earliest  maturity,  June  12,  for  the  focal  date,  we 
should  have  obtained  84  days  for  the  interval  of  time ;  and  since  in 
this  case  the  products  would  represent  the  credit  to  which  the  several 
items  are  entitled  after  June  12,  we  should  add  84  days  to  the  focal 
date,  which  would  give  Sept.  4,  as  before. 

2.  When  is  the  balance  of  the  following  account  due,  per 
average  ? 


.358                                             PERCENTAGE. 

Charles  Derby. 
Dr.                                                                                                                                             Cr. 

1859. 
Jan.     21 
Mar.      5 
"     22 

To  Mdse. 

82 

145 
194 

00 
00 
00 

1859. 
Jan.       1 
Feb.        4 
Mar.     30 

By  cash,                           84      00 
"       "                               40      00 
"       "                               12  f  00 

OPERATION. 


Dr. 


Cr. 


Due. 

Da. 

Items. 

Products. 

Due. 

Da. 

Items. 

Products. 

Jan.    21 

68 

32 

2176 

Jan.      1 

88 

84 

7392 

Mar.      5 

25 

145 

3625 

Feb.      4 

54 

40 

2160 

«      22 

8 

194 

1552 

Mar.   30 

0 

12 

371 

7353 

136 

9552 

136 

7353 

Balance  of  account, 

235 

Balance  of  products. 

2199 

2199  -r  235  =  9  da. ;  Mar.  30  +  9  da.  =  Apr.  8,  Ans. 
ANALYSIS.  We  take  the  latest  maturity,  Mar.  30,  for  the  focal  date, 
and  consequently  the  products  represent  the  interest  due  upon  the 
several  items,  at  that  date.  We  find  the  balance  of  the  items  upon 
the  Dr.  side,  and  the  balance  of  the  products  upon  the  Cr.  side.  The 
debtor  therefore  owes,  on  Mar.  30,  $235,  but  is  entitled  to  such  a  term 
of  interest  on  the  same  as  will  be  equivalent  to  the  interest  on  $2199 
for  1  day,  which  by  division,  is  found  to  be  9  da.  Hence  the  balance 
is  due  Mar.  30+9  da.  =  Apr.  8.  Thus  we  see  that  when  the  balances 
are  on  opposite  sides,  the  interval  of  time  is  counted  from  the  other 
dates.  If  we  take,  in  this  example,  the  earliest  date  for  the  focal  date, 
the  balances  will  both  be  upon  the  Dr.  side,  and  the  interval  of  time 
will  be  97  da.,  which  reckoned  forward  from  the  focal  date,  will  give 
the  equated  time  as  before. 

G2O.    From  these  examples  we  derive  the  following 
RULE.     I.   Find  the  time  when  each  item  of  the  account  is  due, 
and  write  the  dates,  in  two  columns,  on  the  sides  of  the  account  to 
which  they  respectively  belong. 

II.  Use  either  the  earliest  or  the  latest  of  these  dates  as  the  focal 
date  for  both  sides,  ana  Jind  the  products  as  in  the  last  case. 

III.  Divide  the  balance  of  the  products  by  the  balance  of  the 
account ;   the  quotient  will  be  the  interval  of  time,  which  must  be 
reckoned  from  the  focal  date  TOWARD  the  other  dates  when  both 


EQUATION  OF  PAYMENTS. 


359 


balances  are  on  the  same  side  of  the  account,  but  FROM  the  oilier 
dates  when  the  balances  are  on  opposite  sides  of  the  account. 

NOTES.  —  1.  Instead  of  the  products,  we  may  obtain  the  interest,  at  any  per 
cent.,  on  the  several  items  for  the  corresponding  intervals  of  time,  and  divide 
the  balance  of  interest  by  the  interest  on  the  balance  of  the  account  for  1  day  ; 
the  quotient  will  be  the  interval  of  time  to  be  added  to,  or  subtracted  from  the 
focal  date,  according  to  the  rule.  The  time  obtained  will  be  the  same,  at  what- 
ever rate  the  interest  be  computed. 

2,  There  may  be  such  a  combination  of  debits  and  credits,  that  the  equated 
time  will  be  earlier  or  later  than  any  date  of  the  account. 

EXAMPLES   FOR   PRACTICE. 

1.  Required,  the  average  maturity  of  the  following  account. 


Dr. 


A.  Z.  Armour. 


Cr. 


1859. 

11859. 

I 

Feb.      12 

To  Mdse. 

85 

March  15 

By  bal.  old  acc't. 

97   1  36 

«        25 

.4             U 

36 

April     17 

"   cash, 

56  |  00 

April    16 

«             (( 

174 

May      25 

li           U 

25      00 

May      20 

«             « 

94 

June       8 

"   sundries. 

94  1  75 

Dr. 


OPERATION. 


Cr. 


Due. 

Da. 

Items. 

Int 

Due. 

Da. 

Items. 

Int. 

Feb.    12 
25 
April  16 
May    20 

116 
103 
53 
19 

85.75 
36.24 
17496 
94.78 

1.66 
.62      i 
1.55 
.30 

March  15 
April    17 
May      25 
June      8 

85 

52 
14 

97.36 

56.00 
25.00 
94.75 

1.33 

.49 
.06 

Balances, 

391.73 
273.11 

4.13 
1.93 

273.11 

1.93 

118.62 

2.20 

Int  on  $118.62  for  1  da.  =  $,0198. 
2.20-=-.0198=lll  da.;  June  8—111  da.=Feb.  17,1859,  Am. 

ANALYSIS.  Taking  the  latest  maturity,  June  8,  for  the  focal  date, 
we  find  the  interest  of  each  item,  at  6  c/o,  from  its  maturity  to  the 
focal  date ;  then,  taking  the  balance,  we  find  the  interest  due  on  the 
account  to  be  $2.20.  Dividing  this  interest  by  the  interest  on  the 
balance  of  the  items  for  1  day,  we  obtain  111  da.,  the  time  required 
for  the  interest,  $2.20,  to  accrue.  The  average  maturity,  therefore, 
is  June  8  —  111  da.  =  Feb.  17,  1859. 

It  is  evident  that  when  the  balances  occur  on  opposite  sides,  the 
interval  of  time  will  be  reckoned  as  in  the  method  by  products. 


360 


PERCENTAGE. 


2.   What  is  the  balance  of  the  following  account,  and  when  is 
it  due  ? 

Thomas  Lardner. 

Dr.  *  Cr. 


1800. 

I 

1860. 

March  1 

To  Sundries, 

436 

00 

March  25 

By  draft,  at  60  da. 

400 

00 

April  12 

"   Mdse. 

548 

00 

April      6 

"      «          30  " 

650 

00 

July    16 

it      a 

312 

00 

June    20 

"   cash, 

200 

00 

Sept.   14 

"       " 

536 

00 

Aug.       3 

«      « 

84 

00 

Ans.  Balance,  $498;  due  June  22,  1860. 

3.  When  shall  a  draft  for  the  settlement  of  the  following  ac- 
count be  made  payable  ? 

David  Sanford. 


Dr 


Cr. 


1859. 

1859. 

Jan.        1 

To  Mdse.  on  3  mo. 

54 

36 

April    1 

By  cash. 

50 

00 

Feb.       12 

'•       "       "  2     •' 

28 

45 

May    16 

••   draft,  at  30  da. 

30 

00 

March  16 

"    Sundries, 

95 

75 

June  12 

«             H 

128 

00 

June     25 

"   Mdse. 

26 

32 

"      20 

"  cash, 

150 

00 

Ans.  Aug.  28,  1859. 


2>r. 


Oliver  Wainwright. 


Cr. 


1858. 

1858. 

• 

Jan.        1 

To  Mdse. 

36 

72 

Jan.      10 

By  cash, 

98      72 

Feb.         1 

a      a 

48 

25 

21 

tt      n, 

25  1  84 

March  17 

ii      « 

72  '  36 

March  23 

"   sundries, 

15  1  17 

April      1 

«      (i 

98 

48 

April       6 

"        " 

S  1  96 

If  the  above  account  were  settled  April  6,  1858,  by  draft  on 
time,  how  many  days'  credit  should  be  given  ?          Ans.  20  da. 

5.  I  owe  $1000  due  Apr.  25.    If  I  pay  $560  Apr.  1,  and  $324 
Apr.  21,  when,  in  equity,  should  I  pay  the  balance  ? 

Ans.  Aug.  30. 

NOTE. — Make  the  $1000  the  Dr.  side  of  an  account,  and  the  payments  the  Cr. 
«ide,  and  then  average. 

6.  A  man  owes  $684,  payable  Aug.  12,  and  $468,  payable  Oct. 
15.     If  he  pay  $839;  Aug.  1,  what  will  be  the  equated  time  for 
the  payment  of  the  balance  ?  Ans.   Dec.  15. 

7 .  A  man  holds  3  notes,  the  first  for  $500,  due  March  1,  the 
second  for  $800,  due  June  1,  and  the  third  for  $600,  due  Aug.  1. 
He  wishes  to  exchange  them  for  two  others,  one  of  which  shall 
be  for  $1000,  payable  Apr.  1 ;  what  shall  be  the  face  and  when 
the  maturity  of  the  other  ? 

Ans.  Face.  $900 ;  maturity,  July  28. 


EQUATION  OF  PAYMENTS. 


361 


8.  A  owes  $500,  due  Apr  12,  and  $1000,  due  Sept.  20,  and 
wishes  to  discharge  the  obligation  by  two  equal  payments,  made 
at  an  interval  of  60  days ;  when  must  the  two  payments  be  made  ? 

Ana,   1st,  June  28;  2d,  Aug.  27. 

9.  When  shall  a  note  be  made  payable,  to  balance  the  following 

account  ? 

James  Tyler. 


Dr. 


CP. 


1859. 

1859. 

June  12 

To  Mdse.  on  3  mo. 

530 

84 

Sept.  14 

By  cash, 

436 

00 

"      20 

u           u      *.t      tt 

236 

48 

"    25 

*.      u 

320 

00 

<;      30 

it          t:      «      « 

739 

56 

Oct.     3 

«      « 

660 

00 

July      5 

«           ft      it      « 

273 

44 

"    17 

«        u 

370 

00 

"      16 

«          tt     <;     it 

194  |  78 

Nov.  16 

«      « 

840 

00 

"      29 

ti          «     <(     tt 

536  I  42 

1       «•    24 

"      " 

560 

00 

10.  I  received  goods  from  a  wholesale  firm  in  New  York,  in 
parcels,  as  per  bills  received,  namely :  Apr.  1,  a  bill  for  $536.78 ; 
May  16,  $2156.94;  June  12,  $843.75;  July  12,  $594.37;  Sept 
18,  $856.48.  In  part  payment,  I  remitted  cash  as  follows :  June 
3,  $500;  July  1,  $1000;  Nov.  1,  $1500.  When  is  the  balance 
payable,  allowing  credit  of  2  months  for  the  merchandise  ? 

Ans.  July  23. 

ACCOUNT    SALES. 

G21.  An  Account  Sales  is  an  account  rendered  by  a  commis- 
sion merchant  of  goods  sold  on  account  of  a  consignor,  and  con- 
tains a  statement  of  the  sales,  the  attendant  charges,  and  the  net 
proceeds  due  the  owner. 

G22*  Guaranty  is  a  charge  made  in  addition  to  commission, 
for  securing  the  owner  against  the  risk  of  non-payment,  in  case  of 
goods  sold  on  credit 

G23.  Storage  is  a  charge  made  for  keeping  the  goods,  and 
may  be  reckoned  by  the  week  or  month,  on  each  article  or  piece. 

G24L.  Primage  is  an  allowance  paid  by  a  shipper  or  consignor 
of  goods  to  the  master  and  sailors  of  a  vessel,  for  loading  it 

G^«5o    A    commission  merchant   having  sold   a  shipment  of 
goods  by  parts  at  different  times,  and  on  various  terms,  makes  a 
final  settlement  by  deducting  all  charges,  and  accrediting  the  owner 
with  the  net  proceeds.     It  is  evident,  therefore, 
31 


362 


PERCENTAGE. 


I.  That  commission  and  guaranty  should  be  accredited  to  the 
agent  at  the  average  maturity  of  the  sales. 

II.  That  the  net  proceeds  should  be   accredited   to  the  con- 
signor at  the  average  maturity  of  the  entire  account. 

Hence  the  following 

RULE.  I.  To  compute  the  storage.  —  Multiply  each  article  or 
parcel  by  the  time  it  is  in  store,  and  multiply  the  sum  of  the  pro- 
ducts by  the  rate  per  unit;  the  result  will  be  the  storage. 

II.  To  find  when  the  net  proceeds  are  due. — Average  the  sales 
alone,  and  the  result  will  be  the  date  to  be  given  to  the  commission 
and  guaranty  ;  then  make  the  sales  the  Or.  side,  and  the  charges 
the  Dr.  side,  and  average  the  entire  account  by  a  compound  equation. 

NOTE.  —  In  averaging,  either  the  product  method  or  the  interest  method  may 
be  used. 

EXAMPLES    FOR   PRACTICE. 

1.  Account  sales  of  100  pipes  of  gin,  received  per  ship  Hispan- 
iola,  from  Havana,  on  a|c.  of  Tyler,  Jones  &  Co. 


I860. 
April    15 

Sold  39  Pipe*"    4160  gal  @  $105    OD  30 

4368 

00 

May        5 

"    40       "        5240     "     (3),     1  0'2,  cash,. 

53-4 

80 

<•'•    28      "       3650     "     @     1  OJ      " 

3650 

00 

loo    «                              

13362 

80 

April      1 
"         1 

CHARGES. 
To  Freight  and  Primage,  

$136.76 
48  54 

3207.07 

June     28 

"    StorW  from  April  1,  viz.  : 
On  32  Pipes,  2  wks  64  wks. 
"    40      "        5     "      ...  200     « 
«    23      "     13    "      ...  364    " 

100      "      cq  nil  to      628    "  @ 

G  c                      37  68 

*'•   Commission  on  $13362  FO  at  '2%  % 

33407 

"   Girinnty  on  $4368  at  2V£  % 

10920 

3873 

32 

What  are  the  net  proceeds  of  the  above  account,  and  when  due  ? 
Ans.  Net  proceeds,  $9489.48 ;  due,  May  20,  1860. 

NOTE.— The  time  for  which  storage  is  charged  on  each  part  of  the  shipment 
is  the  internal,  reduced  to  weeks,  between  Apr.  1,  when  the  pipes  were  received 
into  store,  anO  the  date  of  sale.  Every  fraction  of  a  week  is  reckoned  a  full  week. 

2.  A  commission  merchant  in  Boston  received  into  his  store  on 
May  1,  1859,  1000  bbl.  of  flour,  paying  as  charges  on  the  same 


EQUATION  OF  PAYMENTS. 


day,  freight  $175.48,  cartage  $56.25,  and  cooperage  $8.37.  He 
sold  out  the  shipment  as  follows:  June  3,  200  bbl.  @  $6.25; 
June  30,  350  bbl.  @  $6.50;  July  29,  400  bbl.  @  $6.12  J;  Aug. 
6,  50  bbl.  @  $6.00.  Kequired  the  net  proceeds,  and  the  date 
when  they  shall  be  accredited  to  the  owner,  allowing  commission 
at  3'i  fa)  and  storage  at  2  cents  per  week  per  bbl. 

Ans.  Net  proceeds,  $5614.28 ;  due,  July  10. 


SETTLEMENT   OF   ACCOUNTS    CURRENT. 

To  find  the  cash  balance  of  an  account  current, 
at  any  given  date. 

/.  Burns  in  account  current  with  Tyler  &  Co. 


Dr. 


Or. 


I860. 

1860. 

Feb.      25 

To  Mdse.  on  3  mo. 

360 

75 

March  1 

By  cash  on  ncct. 

250 

00 

March  20 

3 

240 

50 

April  20 

'•  accept,  at  30  da. 

300 

00 

April    26 

«      "       "  3    « 

875 

24 

June   12 

"  Sundries. 

375 

CO 

June     24 

«      K       u   2    " 

235 

25 

«       27 

"  cash  on  acct. 

4UO 

00 

Required  the  cash  value  of  the  above  account,  July  1,  1860, 
interest  at  6      . 


OPERATION. 


Dr. 


Cr. 


Due. 

Da. 

Items.        Int. 

Cash  ral 

Due. 

Da. 

Items.        Int. 

Cash  val. 

May     25 
June   20 
July    26 
Aug.    24 

37 
11 
25 
54 

360.75  +  2.22 
240.56  -(-    .44 
87  0.24—  3.05 
235.25—2.12 

362.97 
241.00 
871.59 
233.13 

March  1 

May     20 
June  12 
"       —7 

122 
42 
19 
4 

250.00  +  5  OS 
300.00  +  210 
o7-"..00  +  1.19 
400.00  +    .27 

255.08 
302.10 
376.15) 
400.27 

1708.69 

1333.64 

$1708.69  —$1333.64  =  $375.05.  Ans. 

ANALYSIS.  For  either  side  of  the  account  we  write  the  dates  at 
which  the  several  items  are  due,  and  the  days  intervening  between 
these  dates  and  the  day  of  settlement,  July  1.  We  then  compute  the 
interest  on  each  item  for  the  corresponding  interval  of  time,  and  add 
it  to  the  item  if  the  maturity  is  before  July  1,  and  subtract  it  from 
the  item  if  the  maturity  is  after  July  1 ;  the  results  must  be  the  cash 
values  of  the  several  items  on  July  1.  Adding  the  two  columns  of 
cash  values,  and  subtracting  the  less  sum  from  the  greater,  we  have 
$375.05.  the  cash  balance  required.  Hence  the 


364 


PARTNERSHIP. 


RULE.     I.   Find  the  number  of  days  intervening  between  each 
maturity  and  the  day  of  settlement. 

II.  Compute  the   interest  on  each  item  for  t\ie  corresponding 
interval  of  time  ;  add  the  interest  to  the  item  if  the  maturity  is 
before  the  day  of  settlement,  and  subtract  it  from  the  item  if  the 
maturity  is  after  the  day  of  settlement ;  the  results  will  be  the  cash 
Values  of  the  several  items. 

III.  Add  each  column  of  cash  values,  and  the  difference  of  the 
amounts  will  be  the  cash  balance  required. 

EXAMPLES    FOR   PRACTICE. 

1.  Find  the  cash  balance  of  the  following  account  for  June  1, 
1858,  interest  at  6  per  cent.  ? 

Alvan  Parke. 


Dr. 


Cr. 


1858. 

1858. 

Jan.      12 

To  check, 

500 

36 

Jan.       1 

By  bal.  from  old  acct. 

536 

72 

26 

«      (i 

250 

48 

Feb.       3 

«   cash, 

486 

57 

Fob.       13 

a       u 

400 

00 

March  26 

«      « 

1200 

78 

March  16 

a        (t 

750 

00 

April    20 

«      i. 

756 

36 

April    25 

it       (i 

200 

00 

May     12 

c.      ti 

248 

79 

Ans.  $1106.67. 

2.  What  is  the  cash  balance  of  the  following  account  on  Dec. 
31,  at  7  per  cent.  ? 

James  Hanson. 

Dr.  Cr. 


1859. 

ft59. 

Sept.    3 
Oct.      2 

To  Sundries, 
"  Mdse.  on  3.  mo. 

478 
256 

36 
37 

Sept.  17 
"      20 

By  Sundries, 
"  cash  on  acct. 

96 

200 

54 
00 

•'     21 

«      .;       u  3    « 

375 

26 

Oct.      3 

((       <:              t( 

325 

00 

Nov.  12 

c.         li          ti   3      (( 

80 

00 

Nov.   17 

«      i:             (l 

50 

no 

Dec.    15 

"  Sundries, 

148 

76 

Dec.    27 

'•      '•             " 

84 

00 

PARTNERSHIP. 

Partnership  is  a  relation  established  between  two  or 
more  persons  in  trade,  by  which  they  agree  to  share  the  profits 
and  losses  of  business  according  to  the  amount  of  capital  furnished 
by  each,  and  the  time  it  is  employed. 

The  Partners  are  the  individuals  thus  associated. 


NOTE.  —  The  terms  Capital  or  Stock,  Dividend,  and  Assessment,  have  the  same 
lignification  in  Partnership  as  in  Stocks. 


PARTNERSHIP.  365 

CASE  I 

629.  To  find  each  partner's  share  of  the  profit  or 
loss,  when  their  capital  is  employed  for  equal  periods 
of  time. 

1.  A  and  B  engage  in  trade;  A  furnishes  $500,  and  B  $700  as 
capital ;  they  gain  $96 ;  what  is  each  man's  share  ? 

OPERATION.  ^  ANALYSIS.  The  whole 

$  500  amount  of  capital  em- 

$  700  ployed  is  $500  +  $700 

$1200,  whole  stock.  =$1200 ;  hence,  A  fur- 

J LOO  =   5    A's  part  of  the  stock.  nishes  TS%  =ir°f  the 

_7j)0   _  jf   B's     "     "     "     "  capital,  and  B  furnishes 

$9(Px  T<C  =  $40  A's  share  of  the  gain.      ™  =  A  of  the  capi- 
$96  x   7;  =  $56?  B'S     «     "     "     "         taL      And  smce  each 

man's  share  of  the  pro- 
fit or  loss  will  have  the  same  ratio  to  the  whole  profit  or  loss  as  his 
part  of  the  capital  has  to  the  whole  capital,  A  will  have  fV,  of  the 
$96,  and  B  T7-j  of  the  $96,  for  their  respective  shares  of  the  profits. 

We  may  also  regard  the  whole  capital  as  the  first  cause,  and  each 
man's  share  of  the  capital  as  the  second  cause,  the  whole  profit  or  loss 
as  the  first  effect,  and  each  man's  share  of  the  profit  or  loss  as  the 
second  effect,  and  solve  by  proportion  thus : 

1st  cause.  2d  cause.     1st  effect.      2d  effect. 

$1200     :     $500  =  $96     :     ( ? )  =  $40,  A's  gain, 
$1200     :     $700  =  $96     :     (?)  =  $56,B's     « 
Hence  we  have  the  following 

RULE  Multiply  the  whole  profit  or  loss  by  the  ratio  of  the 
whole  capital  to  each  man's  share  of  the  capital.  Or, 

The  whole  capital  is  to  each  mans  share  of  the  capital  as  the 
whole  profit  or  loss  is  to  each  man's  share  of  the  profit  or  loss. 

EXAMPLES   FOR   PRACTICE. 

1.  Three  men  engage  in  trade;  A  puts  in  $6470,  B  $3780,  and 
C  $9860,  and  they  gain  $7890.  What  is  each  partner's  share  of  the 
profit?  Ans.  A's,  $2538.453;  B's,  $1483.053;  C's,  $3868.493. 

2.,   B  and  C  buy  pork  to  the  amount  of  $1847.50,  of  which  B 
pays  $739,  and  C  the  remainder.     They  gain  $375 ;  what  is  each 
one's  share  of  the  gain  ? 
31* 


366  PARTNERSHIP. 

3.  A,  B,  and  C  form  a  company  for  the  manufacture  of  woolen 
cloths.     A  puts  in  $10000,  B  $12800,  and  C  $3200.    C  is  allowed 
$1500  a  year  for  personal  attention  to  the  business;    their  ex- 
penses for  labor,  clerk  hire,  and  other  incidentals  for  1  year  are 
$3400,  and  their  receipts  during  the  same  time  are  $9400.    What 
is  A's,  B's,  and  C's  income  respectively  from  the  business? 

4.  Four  persons  rent  a  farm  of  115  A.  32  P,  at  $3.75  an  acre. 
A  puts  on  144,  B  160,  C  192,  and  D  324  sheep;  how  much  rent 
ought  each  to  pay  ? 

5.  Three  persons  gain  $2640,  of  which  B  is  to  have  $6  as  often 
as  C  $4,  and  as  often  as  D  $2  ;  how  much  is  each  one's  share? 

6.  Six  persons  are  to  share  among  them  $6300;  A  is  to  have 
-^  of  it,  B  -J-,  C  f ,  D  is  to  have  as  much  as  A  and  C  together,  and 
the  remainder  is  to  be  divided  between  E  and  F  in  the  ratio  of 
3  to  5.     How  much  does  each  receive  ? 

Ans.  A,  $900;    B,  $1260;    C,  $1400; 
D,  $2300  ;  E,  $165 ;      F,  $275. 

7.  Two  persons  find  a  watch  worth  $90,  and  agree  to  divide  the 
value  of -it  in  the  ratio  of  J-  to  -|;  how  much  is  each  one's  share? 

NOTE.— If  the  fractions  be  reduced  to  a  common  denominator,  they  will  be  to  each 
other  as  their  numerators,  (418,  HI). 

8.  A  father  divides  his  estate  worth  $5463.80  between  his  two 
sons,  giving  the  elder  -J-  more  than  the  younger;  how  much  is  each 
son's  share?  Ans.  Elder,  $2892.60;  younger,  $2571.20. 

9.  Three  men  trade  in   company.     A  furnishes  $8000,  and  B 
$1 2000.   Their  gain  is  $1680,  of  which  C's  share  is  $840  ;  required, 
C's  stock,  and  A's  and  B's  gain.  Ans.  C's  stock,  $20000. 

10.  Four  persons  engage  in  the  lumber  trade,  and  invest  jointly 
$22500;    at  the  expiration  of  a  certain  time,  A's  share  of  the 
gain  is  $2000,  B's  $2800.75,  C's  $1685.25,  and  D's  $1014;  how 
much  capital  did  each  put  in  ?  '    Ans.  D  put  in  $3042. 

11.  A  legacy  of  $30000  was  left  to  four  heirs  in  the  propor- 
tion  of  ^-,  -|,  -J-,  and  -J-,  respectively ;  how  much  was  the  share  of 
each  ? 

12.  Three  men  purchase  a  piece  of  land  for  $1200,  of  which 
sum  C  pays  $500.     They  sell  it  so  as  to  gain  a  certain  sum  of 


PARTNERSHIP.  367 

which  A  takes  $71.27,  and  B  $142.54;  how  much  do  A  and  B 
pay,  and  what  is  C's  share  of  the  gain  ?    Ans.  C's  gain,  $152.  72{. 

13.  Three  persons  enter  into  partnership  for  the  manufacture 
of  coal  oil,  with  a  joint  capital  of  $18840.     A  puts  in  83  as  often 
as  B  puts  in  $5,  and  as  often  as  C  puts  in  $7.     Their  annual  gain 
is  equal  to  C's  stock;  how  much  is  each  partner's  gain? 

14.  A,  B,  and  C  are  employed  to  do  a  piece  of  work  for  §26  45. 
A  and  B  together  are  supposed  to  do  |  of  the  work,  A  and  C  -y9^ 
and  B  and  C  AJ,  and  are  paid  proportionally;  how  much  must 
each  receive?  *  Ans.  A,  $11.50;  B,  $5  75;  C,  $9.20. 

CASE   II. 

63O.  To  find  each  partner's  share  of  the  profit  or  loss 
when  their  capital  is  employed  for  unequal  periods  of 
time. 

It  is  evident  that  the  respective  shares  of  profit  and  loss  will 
depend  equally  upon  two  conditions,  viz.  :  the,  amount  of  capital 
invested  by  each,  and  the  time  it  is  employed.  Hence  they  will 
be  proportional  to  the  products  of  these  two  elements. 

1.  Two  men  form  a  partnership  ;  A  puts  in  $320  for  5  months, 
and  B  $400  for  6  months.  They  lose  $140;  what  is  each  man's 
share  of  the  loss  ? 

OPERATION. 

$320  X  5  =  $1600,  A's  capital  for  1  mo 
6400  x  6  =  $2400.  B's      «       «      « 

$4000,  entire  "        "       « 
=  |,  A's  share  in  the  partnership 


$140  x  -|  •=  $56,  A'=  loss 
$140  x  f  =  884,  B's  loss. 

ANALYSIS  The  use  of  $320  for  5  months  is  the  same  as  the  use  of 
5  times  $320,  or  $1600,  for  1  month  ;  and  the  use  of  $400  for  6  months 
is  the  same  as  the  use  of  6  times  $400,  or  §2400,  for  1  month  ;  hence 
the  use  oi  the  entire  capital  is  the  same  as  the  use  of  $1600  -f-  $2400 
=  $4000  for  1  month.  A's  interest  in  the  partnership  is  therefore 

2  =  $55  .     g^ 


368  PARTNERSHIP. 

B's  interest  in  the  partnership  is  f ^§£  =  f ,  and  he  will  suffer  |  of  the 

loss,  or  $140  X  §  •=  $84. 

We  may  also  solve  by  proportion,  the  causes  being  compounded  of 

the  two  elements,  capital  and  time  •  thus : 

$4000  :  $1600  =  $140  :  (?)  =  $56,  A's  loss 
$4000  :  $2400  =  $140  :  (?)  =  $84;  B's  loss. 

Hence  the  following 

RULE,  Multiply  each  man's  capital  by  the  time  it  is  employed 
in  trade,  and  add  the  products.  Then  multiply  the  entire  profit  or 
loss  by  the  ratio  of  eacji  product  to  the  sum  of  the  products  ;  the 
results  will  be  the  respective  shares  of  profit  or  loss  of  each  part- 
ner. Or, 

Multiply  each  man's  capital  by  the  time  it  is  employed  in  trade, 
and  regard  each  product  as  his  capital,  and  the  sum  of  the  pro~ 
ducts  as  the  entire  capital,  and  solve  by  proportion,  as  in  Case  L 

EXAMPLES   FOR   PRACTICE. 

1.  A,  B,  ana  C  enter  into  partnership.     A  pnts  in  $357  for  5 
months,  B  $371   for  7  months,  and  C  $154  for  11  months,  and 
they  gain  $347.20 ;  how  much  is  each  one's  share  ? 

Ans.  A's,  $102  ;  B's  $148.40  ;  C's  $96.80. 

2.  Three  men  hire  a  pasture  for  $55.50.     A  put  in  5  cows,  12 
weeks;  B,  4  cows,  10  weeks;  and  C,  6  cows,  8  weeks;  how  much 
ought  each  to  pay  ?  Ans.  A  $22.50  ;  B  $15 ;  C  $18. 

3.  B  commenced  business  with  a  capital  of  $15000.  Three 
months  afterward  C  entered  into  partnership  with  him,  and  put 
in  125  acres  of  land.  At  the  close  of  the  year  their  profits  were 
$4500,  of  which  C  was  entitled  to  $1800;  what  was  the  value  of 
the  land  per  acre  ? 

4.  A  and  B  engaged  in  trade.  A  put  in  $4200  at  first,  and  9 
months  afterward  $200  more.  B  put  in  at  first  $1500,  and  at  the 
end  of  6  months  took  out  $500.  At  the  end  of  16  months  their 
gain  was  $772.20 ;  how  much  is  the  share  of  each  ? 

5.  Four  companies  of  men  worked  on  a  railroad.  In  the  first 
company  there  were  30  men  who  worked  12  days,  9  hours  a  day; 
in  the  second,  there  were  32  men  who  worked  15  days,  10  hours 


PARTNERSHIP.  359 

a  day;  in  the  third,  there  were  28  men  who  worked  18  days,  11 
hours  a  day;  and  in  'the  fourth,  there  were  20  men  who  worked 
15  days,  12  hours  a  day.  The  entire  amount  paid  to  all  the  com- 
panies was  $1500;  how  much  were  the  wages  of  each  company? 

6.  A  and  B  are  partners.     A's  capital  is  to  B's  as  5  to  8 ;  at 
the  end  of  4  months  A  withdraws  £  of  his  capital,  and  B  f  of  his ; 
at  the  end  of  the  year  their  whole  gain  is  $4000 ;  how  much  be- 
longs to  each  ?  An*.  A,  $1714^ ,  B,  $2285$. 

7.  B,  C,  and  D  form  a  manufacturing  company,  with  capitals 
of  $15800,  $25000,  and  $30000  respectively.     After  4  months  B 
draws  out  $1200,  and  in  2  months  more  he  draws  out  $1500  more, 
and  4  months  afterward  puts  in  $1000.     C  draws  out  $2000  at 
the  end  of  6  months,  and  $1500  more  4  months  afterward,  and  a 
month  later  puts  in   $800.     D  puts  in   $1800  at  the  end  of  7 
months,  and  3  months  after  draws  out  $5000.     If  their  gain  at 
the  end  of  18  months  be  $15000,  how  much  should  each  receive? 

AH*.  B,  $3228.07;  C,  $5258.15;  D,  $6513.78. 

8.  The  joint  stock  of  a  company  was  $5400,  which  was  doubled 
at  the  end  of  the  year.     A  put  J  for  f  of  a  year,  B  f  for  £  a  year, 
and  C  the  remainder  for  one  year.     How  much  is  each  one's  share 
of  the  entire  stock  at  the  end  of  the  year  ? 

9.  Three  men  engage  in  merchandising.     A's  money  was  in 
10  months,  for  which  he  received  $456  of  the  profits;  B's  was  in 
8  months,  for  whi&h  he  received  $343.20  of  the  profits;  arid  C's 
was  in  12  months,  for  which  he  received  $750  of  the  profits.  Their 
whole  capital  invested  was  $14345 ;  how  much  was  the  capital  of 
each?  Ans.  A's,  34332;  B's,  $4075.50;  C's,  $5937.50. 

10.  Three  men  take  an  interest  in  a  coal  mine.     B  invests  his 
capital  for  4  months,  and  claims  ^  of  the  profits ;  C's  capital  is  in 
8  months ;  and  D  invests  $6000  for  6  months,  and  claims  |  of  the 
profits ;  how  much  did  B  and  C  put  in  ? 

J  11.  A,  B,  and  C  engage  in  manufacturing  shoes.  A  puts  in 
$1920  for  6  months;  B,  a  sum  not  specified  for  12  months;  and 
C,  $1280  for  a  time  not  specified.  A  received  $2400  for  his  stock 
and  profits,  B  $4800  for  his,  and  C  $2080  for  his.  Require^ 
B's  stock,  and  C's  time  ? 

y 


870  ALLIGATION. 


ALLIGATION. 

i 

O31.  Alligation  treats  of  mixing  or  compounding  two  or 
more  ingredients  of  different  values  or  qualities. 

632.  The  Mean  Price  or  Quality  is  the  average  price  or 
quality  of  the  ingredients,  or  the  price  or  quality  of  a  unit  of  the 
mixture. 

CASE   I. 

633.  To  find  the  mean  price  or  quality  of  a  mixture, 
when  the  quantity  and  price  of  the  several  ingredient* 
are  given. 

NOTE. — The  process  of  finding  the  mean  or  average  price  of  several  ingredi- 
ents is  called  Alligation  Medial. 

1.  A  produce  dealer  mixed  together  84  bushels  of  oats  worth 
$.30  a  bushel,  60  bushels  of  oats  worth  $.38  a  bushel,  and  56 
bushels  of  oats  worth  $  40  a  bushel ;  required,  the  mean  price. 

OPERATION.  ANALYSIS.     The  worth  of  84 

$.30  x  84  =  $25.20  bushels  @  $.30  is  $25.20,  of 

.38  X  60  =    22.80  CO  bushels  @  $.38  is  $22.80, 

.40  x  56  =    22.4.0  and  of  56  bushels  @  $.40  is 

200     ")  $70  40  $22.40 ;  and  we  have  in  the 

whole  compound  84  -f-  60  +  56 

$.3520,  Ans.         =  200  bushels,  worth  $25.20+ 

$22.80  +  $22.40  =  $70.40.     One  bushel  of  the  mixture  is  therefore 

worth  $70.40  -f-  200  —  $.352.     Hence  the  following 

RULE.  Find  the  entire  cost  or  value  of  the  ingredients,  and 
divide  it  by  the  sum  of  the  simples. 

EXAMPLES   FOR   PRACTICE. 

1.  A  grocer  mixed  4  Ib.  of  tea  at  $.60  with  3  Ib.  at  $.70,  1  ib. 
at  $1.10,  and  2  Ib.  at  $1.20;  how  much  is  1  Ib.  of  the  mixture 
worth?  Ans.  $.80. 

2.  A  dealer  in  liquors  would  mix  14  gal.  of  water  with  12  gal. 
of  wine  at  $.75,  24  gal.  at  $.90,  and  16  gal.  at  $1.10;  how  muck 
is  a  gallon  of  tho  mixture  worth  ?  Ans.  $.73^. 


ALLIGATION.  371 

3.  If  3  lb.  6  oz.  of  gold  23  carats  fine  be  compounded  with 
4  Ib.  8  oz.  21  carats,  3  lb.  9  oz.  20  carats,  and  2  lb.  2  oz.  of  alloy, 
what  is  the  fineness  of  the  composition  ?  Ans.  18  carats. 

4.  A  grain  dealer  mixes  15  bu.  of  wheat  at  $1.20  with  5  bu. 
at  $1.10,  5  bu.  at  $.90,  and  10  bu.  at  $.70 ;  what  will  be  his  gain 
per  bushel  if  he  sell  the  compound  at  $1.25. 

5.  A  merchant  sold  17  lb.  of  sugar  at  5  cts.  a  pound,  51  lb.  at 
8  cts.,  68  lb.  at  10  cts.,  17  lb.  at  12  cts.,  and  thereby  gained  on 
the  whole  33 £  per  cent.;  how  much  was  the  average  cost  per 
pound  ? 

6.  A  drover  bought  42  sheep  at  $2.70  per  head,  48  at  $2.85, 
and  65  at  $3.24 ;  at  what  average  price  per  head  must  he  sell 
them  to  gain  20  per  cent.?  Ans.  $3.567^f. 

7.  A  surveyor  took  10  sets  of  observations  with  an  instrument, 
for  the  measurement  of  an  angle,  with  the  following  results :  1st, 
36°  17'  25.4";  2d,  36°  17'  24.5";  3d,  36°  17'  27.8";  4th,  36°  17' 
26.9";  5th,  36°  17'  25.4";  6th,  36°  17'  24.7";  7th,  36°  17'  24.2"; 
8th,  36°  17'  26.3";  9th,  36°  17'  25.8";  10th,  36°  17'  26.7".  What 
is  the  average  of  these  measurements  ?       Ans.  36°  17'  25.77". 

8.  Three  trials  were  made  with  chronometers  to  determine  the 
difference  of  time  between  two  places;  the  first  trial  gave  37  min. 
54.16  sec.,  the  second  37  min.  55.56  sec.,  and  the  third  37  min. 
54.82  sec.     Owing  to  the  favorable  conditions  of  the  third  trial, 
it  is  entitled  to  twice  the  degree  of  reliance  to  be  placed  upon 
either  of  the  others ;  what  should  be  taken  as  the  difference  of 
longitude  between  the  two  places,  according  to  these  observations? 

Ans.  9°  28' 42.6". 

CASE   II. 

634.  To  find  the  proportional  quantity  to  be  used  of 
each  ingredient,  when  the  mean  price  and  the  prices  of 
the  several  simples  are  given. 

NOTE. — The  process  of  finding  the  quantities  to  be  used  in  any  required  mix- 
ture is  commonly  called  Alligation  Alternate. 

I.  A  farmer  would  mix  oats  worth  3  shillings  a  bushel  with 
peas  worth  8  shillings  a  bushel,  to  make  a  compound  worth  5  shil- 
lings a  bushel ;  what  quantities  of  each  may  he  take  ? 


372 


ALLIGATION. 


OPERATION. 


(3    £   3) 

18    I  8J 


Ans. 


ANALYSIS.  If  a  mixture,  in  any  pro- 
portions, of  oats  worth  3  shillings  a 
bushel  and  peas  worth  8  shillings,  be 
priced  at  5  shillings,  there  will  be  a 
gain  on  the  oats,  the  ingredient  worth  less  than  the  mean  price,  and 
a  loss  on  the  peas,  the  ingredient  worth  more  than  the  mean  price ; 
and  if  we  take  such  quantities  of  each  that  the  gain  and  loss  shall 
each  be  1  shilling,  the  unit  of  value,  the  result  will  be  the  required 
mixture.  By  selling  1  bushel  of  oats  worth  3  shillings  for  5  shil- 
lings, there  will  be  a  gain  of  5  —  3  =  2  shillings,  and  to  gain  1  shil- 
ling would  require  £  of  a  bushel ;  hence  we  place  £  opposite  the  3.  By 
gelling  1  bushel  of  peas  worth  8  shillings  for  5  shillings,  there  will 
be  a  loss  of  8  —  5  =  3  shillings,  and  to  lose  1  shilling  will  require  £ 
of  a  bushel ;  hence  we  write  £  opposite  the  8.  Therefore,  £  bushel  of 
oats  to  ^  of  a  bushel  of  peas  are  the  proportional  quantities  for  the 
required  mixture.  It  is  evident  that  the  gain  and  loss  will  be  equal, 
if  we  take  any  number  of  times  these  proportional  terms  for  the  mix- 
ture. We  may  therefore  multiply  the  fractions  \  and  \  by  6,  the  least 
common  multiple  of  their  denominators,  and  obtain  the  integers  3 
and  2  for  the  proportional  terms  (418,111);  that  is,  we  may  take,  for  the 
mixture,  3  bushels  of  oats  to  2  bushels  of  peas. 

2.  What  relative  quantities  of  sugar  at  7  cents,  8  cents,  11  cents, 
and  14  cents  per  pound,  will  produce  a  mixture  worth  10  cents 
per  pound  ? 

OPERATION.  ANALYSIS.      To    preserve    the 

equality  of  gains  and  losses,  we 
must  compare  two  prices  or  sim- 
ples, one  greater  and  one  less  than 
the  mean  rate,  and  treat  each  pair 
or  couplet  as  a  separate  example. 
Thus,  comparing  the  simples  whose 
prices  are  7  cents  and  14  cents,  we 
find  that,  to  gain  1  cent,  £  of  a 
pound  at  7  cents  must  be  taken, 
and,  to  lose  1  cent,  \  of  a  pound  at 
14  cents  must  be  taken  ;  and  com- 
paring the  simples  the  prices  of 
which  are  8  cents  and  11  cents,  we 
pound  at  8  cents  must  be  taken  to  gain  1  cent,  and  1  pound 


10 


10 


find  that 


1 

2 

3 

4 

5 

7 

7 

~± 

4 

8 

\ 

1 

1 

11 

1 

2 

2 

14 

i 

3 

3 

Or, 

1 

2 

3 

f    7 

7 

1 

J     8 

* 

4 

I11 

i 

3 

114 

\ 

2 

11  cents  must  be  taken  to  lose  1  cent.    These  proportional  terms  are 


ALLIGATION.  373 

•written  in  columns  1  and  2  We  now  reduce  these  couplets  separately 
to  integers,  as  in  the  last  example,  writing  the  results  in  columns  3 
and  4 ;  and  arranging  all  the  terms  in  column  5>  we  have  4,  1,  2,  and 
3  for  the  proportional  quantities  required.  If  we  compare  the  prices 
7  and  11  fcr  the  first  couplet,  and  the  prices  8  and  14  for  the  second 
couplet,  as  in  the  second  operation,  we  shall  obtain  1 ,  4,  3,,,  and  2  for 
the  proportional  terms. 

It  will  be  seen  that  in  comparing  the  simples  of  any  couplet,  one 
of  which  is  greater  and  the  other  less  than  the  mean  rate,  the  pro- 
portional number  finally  obtained  lor  either  term  is  the  difference 
between  the  mean  rate  and  the  other  term.  Thus,  in  comparing  7 
and  14,  the  proportional  number  corresponding  to  the  former  simple 
is  4,  which  is  the  difference  between  14  and  the  mean  rate  10  ;  and 
the  proportional  number  corresponding  to  the  latter  simple  is  3, 
which  is  the  difference  between  7  and  the  mean  rate.  The  same  is 
true  of  every  other  couplet.  Hence,  when  the  simples  and  the  mean 
rate  are  integers,  the  intermediate  steps  taken  to  obtain  the  final  pro- 
portional numbers  as  in  columns  1,  2,  3,  and  4.  may  be  omitted,  and 
the  same  results  readily  found  by  taking  the  difference  between  each 
simple  and  the  mean  rate,  and  placing  it  opposite  the  one  with  which 
it  is  compared. 

From  these  examples  and  analyses  we  derive  the  following 

RULE.  I.  Write  the  prices  or  qualities  of  the  several  ingre- 
dients in  a  column,  and  the  mean  price  or  quality  at  the  left. 

II.  Consider  any  two  prices,  one  of  which  is  less  and  the  other 
greater  than  the  mern  rate,  as  forming  a  couplet ;  find  the  differ- 
ence between  each  of  these  prices  and  the  mean  rate,  and  write  the 
reciprocal  of  each  difference  opposite  the  given  price  in  the  couplet, 
as  one  of  the  proportional  terms.    In  like  manner  form  the  couplets, 
till  all  the  prices  have  been  employed,  writing  each  pair  of  propor~ 
tional  terms  in  a  separate  column. 

III.  If  the  proportional  terms  thus  obtained  are  fractional,  mul' 
tiply  each  pair  by  the  least  common  multiple  of  their  denominators, 
and  carry  these  integral  products  to  a  single  column,  observing  to 
add  any  two  or  more  that  stand  in  the  same  horizontal  line ;  the 
final  results  will  be  the  proportional  quantities  required. 

NOTES. — 1.  If  the  numbers  in  any  couplet  or  column  have  a  common  factor, 
it  may  be  rejected. 

32 


374  ALLIGATION. 

2.  We  may  also  multiply  the  numbers  in  any  couplet  or  column  by  any  ran!- 
tiplier  we  choose,  without  affecting  the  equality  of  the  gains  and  losses,  and 
thus  obtain  an  indefinite  number  of  results,  any  one  of  which  being  taken  will 
give  a  correct  final  result.  \ 

EXAMPLES    FOE   PRACTICE. 

1.  What  quantities  of  flour  worth  $5£,  $6,  and  $7f  per  barrel, 
must  be  sold,  to  realize  an  average  price  of  $6£  per  barrel? 

OPERATION.  ANALYSIS.       Comparing    the 


6*4  6 


II  4 


4| 


I    4 


12 
2      2 


12 
4 


first  price  with  the  third,  we  ob- 
tain the  couplet  5  to  -f ;  and  com- 
paring the  second  price  with  the 


third,  we  obtain  the  couplet  4  to 
§.  Reducing  these  proportional  terms  to  integers,  we  find  that  we 
may  take  4  barrels  of  the  first  kind  with  2  of  the  third,  and  12  of  the 
second  kind  with  2  of  the  third ;  and  these  two  combinations  taken 
together  give  4  of  the  first  kind,  12  of  the  second,  and  4  of  the  third. 
2.  How  much  sugar  worth  5  cts.,  7  cts.,  12  cts.,  and  13  cts.  per 
pound,  will  form  a  mixture  worth  10  cts.  per  pound? 

3  Ib.  of  each  of  the  first  and  third  kinds,  2  Ib. 

of  the  second,  and  5  Ib.  of  the  fourth. 
8.  How  can  wine  worth  $.60  $.90  and  $1.15  per  gallon  be  mixed 
with  water  so  as  to  form  a  mixture  worth  $.75  a  gallon  ? 

.       (  By  taking  3  gal.  of  each  of  the  first  two  kinds  of 
1    wine,  15  gal.  of  the  third,  and  8-gal.  of  water. 

4.  A  farmer  has  3  pieces  of  land  worth  $40,  $60,  and  $80  an 
acre  respectively      How  many  acres  must  he  sell  from  the  dif- 
ferent tracts,  to  realize  an  average  price  of  $62.50  an  acre? 

5.  How  much  wine  worth  $.60,  $.50,  $.42,  $.38,  and  $.30  pe? 
pint,  will  make  a  mixture  worth  $.45  a  pint  ? 

6.  What  relative  quantities  of  silver  |  pure,  |  pure,  and  fy 
pure,  will  make  a  mixture  |  pure  ? 

Ans.   3  Ib.  |  pure,  3  Ib.  |  pure,  and  20  Ib.  -ft  pur3. 

CASE   III. 

O3*>.  When  two  or  more  of  the  quantities  are  re- 
quired to  be  in  a  certain  proportion. 

1-  A  farmer  having  oats  worth  $.30,  com  worth  $.00,  and  wheat 


ALLIGATION.  _          375 

worth  $1.10  per  bushel,  desires  to  form  a  mixture  worth  $.50  pei 
bushel,  which  shall  contain  equal  parts  of  corn  and  wheat ;  in 
what  proportion  shall  the  ingredients  be  taken  ? 


OPERATION.  ANALYSIS.     "VVe  first  obtain 


30 
50  -j  C60 

gillO 


ill!! 


A 


G 


the  proportional  terms  in  col- 
umns 3  and  4,  by  Case  II. 
Now,  it  is  evident  that  the  loss 
and  gain  will  be  equal  if  we 
take  each  couplet,  or  any  mul- 
tiple of  each,  alone;  or  both 

couplets,  or  any  multiples  of  both,  together.  Multiplying  the  terms 
in  column  4  by  2,  we  obtain  the  terms  in  column  5  ;  and  adding  the 
terms  in  columns  3  and  5,  we  obtain  the  terms  in  column  6  ;  that  is, 
the  farmer  takes  7  bushels  of  oats  to  2  of  corn  and  2  of  wheat,  which 
is  the  required  proportion.  Hence  the  following 

RULE.  I.  Compare  the  given  prices,  and  obtain  the  proportional 
terms  by  couplets,  as  in  Case  II. 

II.  Reduce  the  couplets  to  higher  or  lower  terms,  as  may  be  re- 
quired; then  select  the  columns  at  pleasure,  and  combine  them  by 
adding  the  terms  in  the  same  horizontal  line,  till  a  set  of  pro- 
portional terms  is  obtained,  answering  the  required  conditions. 

EXAMPLES    FOR   PRACTICE. 

1.  A  grocer  has  four  kinds  of  molasses,  worth  $.25,  $.50,  $.62, 
and  $.70  per  gallon,  respectively ;  in  what  proportions  may  he  mix 
the  four  kinds,  to  obtain  a  compound  worth  $.58  per  gallon,  using 
equal  parts  of  the  first  two  kinds?  Ans.  4,  4,  8  and  11. 

2.  In  what  proportions  may  we  take  sugars  at  7  cts.,  8  cts.,  13 
cts.,  and  15  cts.,  to  form  a  compound  worth  10  cts.  per  pound,  using 
equal  parts  of  the  first  three  kinds  ?  Ans.  5,  5,  5  and  2. 

8  A  miller  has  oats  at  30  cts.,  corn  at  50  cts.,  and  wheat  at 
100  cts.  per  bushel.  He  desires  to  form  two  mixtures,  each  worth 
70  cts.  per  bushel.  In  the  first  he  would  have  equal  parts  of  oats 
and  corn,  and  in  the  second,  equal  parts  of  corn  and  wheat;  what 
must  be  the  proportional  terms  for  each  mixture  ? 

.        (  For  the  first  mixture,  1,  1  and  2. 
(  For  the  second  mixture,  1,  4  and  4. 


376  ALLIGATION. 

CASE  IV. 

636.  When  the  quantity  of  one  of  the  simples  is 
limited. 

1  A  miller  has  oats  worth  $.28,  corn  worth  $.44,  and  barley- 
worth  $.90  per  bushel.  He  wishes  to  form  a  mixture  worth  $.58 
per  bushel,  and  containing  100  bushels  of  corn.  How  many 
bushels  of  oats  and  barley  may  he  take  ? 

OPERATION.  ANALYSIS.  By  Case 


r28  1 

58  J  44]  | 

(93JJ 


31!, 


A 


7 


7  1140 


5 


6  I  2 


100 
160 


II,  we  find  the  pro- 
portional quantities 
to  be  7  bushels  of 


oats  to  5  of  corn  and 
8  of  barley.  But  as  100  bushels  of  corn,  instead  of  5,  are  required, 
we  must  take  *f  °  =  20  times  each  of  the  other  ingredients,  in  order 
that  the  gain  and  loss  may  be  equal ;  and  we  shall  therefore  have 
7  X  20  =  140  bushels  01  oats,  and  8  X  20  =  160  bushels  of  barley. 
Hence  the  following 

RULE.  Find  the  proportional  quantities  by  Case  II  or  Case 
III.  Divide  the  given  quantity  by  the  proportional  quantity  of 
this  ingredient,  and  multiply  each  of  the  other  proportional  quan- 
tities by  the  quotient  thus  obtained. 

EXAMPLES   FOR   PRACTICE. 

1.  A  dairyman  bought  10  cows  at  $20  a  head ;  how  many  must 
he  buy  at  $16,  $18,  and  $24  a  head,  so  that  the  whole  may  cost 
him  an  average  price  of  $22  a  head  ? 

Ans.  10  at  $16,  10  at  $18,  and  60  at  $24. 

2.  Bought  12  yards  of  cloth  for  $15;  how  many  yards  must  I 
buy  at  $lf,  and  $|  a  yard,  that  the  average  price  of  the  whole 
may  be  $li  ?  Ans.  12  yards  at  $lf  and  16  yards  at  $|. 

3.  How  much  water  will  dilute  9  gal.  2  qt.  1  pt.  of  alcohol  96 
per  cent,  strong  to  84  per  cent.  ?  Ans.  1  gal.  1  qt.  1  pt. 

4.  A  grocer  mixed  teas  worth  $.30,  $.55,  and  $.70  per  pound 
respectively,  forming  a  mixture  worth  $.45  per  pound,  having  equal 
parts  of  the  first  two  kinds,  and  12  Ibs.  of  the  third  kind;  how 
many  pounds  of  each  of  the  first  two  kinds  did  he  take  ? 


ALLIGATIOK.  377 

CASE   V. 

637.  When  the  quantities  of  two  or  more  of  the  in- 
gredients are  limited. 

1.  How  many  bushels  of  rye  at  $1.08,  and  of  wheat  at  $1.44, 
must  be  mixed  with  18  bushels  of  oats  at  $.48,  8  bushels  of  corn 
at  $.52,  and  4  bushels  of  barley  at  $.85,  that  the  mixture  may  be 
worth  $.84  per  bushel  ? 

OPERATION.  ANALYSIS.     Of  the  given 

ciOvxio  CQ«4  quantities   there    are    18  -f 

? 


52  x    8  -         4  16  8  +  4  =  30 

*  "  >  '  mean   or   average   prce  we 

.80  X     4   ==  V 


30     )     $16.2(  We  are  therefore  required  to 

Mean  price  of  the )  *       r±  mix    30    bushels    of   grain 

given  simples      j  worth  $.54  per  bushel,  with 

I  3^  ]  4  !  2   6  I  30 1  rye  at  $1.08,  and  wheat  at 

5        5  1 25  $1.44,  to  make  a  compound 

1   1 '    5  worth  $.84  per  bushel.    Pro- 


54 

84-!  108 


ceeding  as  in  Case  IV,  we 
find  there  will  be  required  25  bushels  of  rye,  and  5  bushels  of  wheat. 
Hence  the  following 

RULE.  Consider  those  ingredients  whose  quantities  and  prices 
are  given  as  forming  a  mixture,  and  find  their  mean  price  ty  Case 
I;  then  consitler  this  mixture  as  a  single  ingredient  ivhose  quantity 
and  price  are  known,  and  fold  the  quantities  of  the  other  ingredients 
ly  Case  IV. 

EXAMPLES   FOR  PRACTICE. 

1.  A  gentleman  bought  7  yards  of  cloth  @  $2.20,  and  7  yards 
@  $2;  how  much  must  he  buy  @  $1.60,  and  @  $1.75  that  the 
average  price  of  the  whole  may  be  $1.80  ? 

2.  How  much  wine,  at  $1.75  a  gallon,  must  be  added  to  60  gal- 
lons at  $1.14,  and  30  gallons  at  $1.26  a  gallon,  so  that  the  mixture 
may  be  worth  $1.57  a  gallon  ?  Ans.  195  gallons. 

3.  A  farmer  has  40  bushels  of  wheat  worth  $2  a  bushel,  and 
70  bushels  of  corn  worth  $£  a  bushel.     How  many  oats  worth  $J 
a  bushel  must  he  mix  with  the  wheat  and  corn,  to  make  the  mix- 
ture worth  $1  a  bushel  ?  Ans.  6|  bushels. 

32* 


378 


ALLIGATION. 


CASE   VI. 

638.  When  the  quantity  of  the  whole  compound  is 
limited. 

1.  A  tradesman  has  three  kinds  of  tea  rated  at  $.30,  8.45,  and 
6.60  per  pound,  respectively;  what  quantities  of  each  should  he 
take  to  form  a  mixture  of  72  pounds,  worth  $.40  per  pound? 

OPERATION.  ANALYSIS.    By  Case  II, 


2 


3 

4 

5 

6 

2 

1 

3 

36 

2 

2 

24 

1 

1 

12 

6 

~T2 

we  find  the  proportional 
quantities  to  form  the 
mixture  to  be  3  Ib.  at 
$.30,  2  Ib.  at  $.45,  and 

GO    ^V  1          1    12  1   lb-    at   $-6()-      Adding 

these  proportional  quanti- 
ties, we  find  that  they 
would  form  a  mixture  of  6  pounds.  And  since  the  required  mixture 
is  7g2  =  12  times  6  pounds,  we  multiply  each  of  the  proportional  terms 
by  12,  and  obtain  for  the  required  quantities,  36  Ib.  at  $.30,  24  Ib.  at 
$.45,  and  1 2  Ib.  at  $.60.  Hence  the  following 

RULE.  Find  the  proportional  numbers  as  in  Case  II or  Case 
III.  Divide  the  given  quantity  by  the  sum  of  the  proportional 
quantities,  and  multiply  each  of  the  proportional  quantities  by  the 
quotient  thus  obtained. 

EXAMPLES   FOR   PRACTICE. 

1.  A  grocer  has  coffee  worth  8  cts.,  16  cts.,  and  24  cts.  per 
pound  respectively ;  how  much  of  each  kind  must  he  use,  to  fill  a 
cask  holding  240  Ib,  that  shall  be  worth  20  cts.  a  pound  ? 

Ans.  40  Ib.  at  8  cts  ,  40  Ib.  at  16  cts.,  and  160  Ib.  at  24  cts. 

2.  A  man  bought  calves,  sheep,  and  lambs,  154  in  all,  for  $154. 
He  paid  $3£  for  each  calf,  $1*  for  each  sheep,  and  $£  for  each 
lamb ;  how  many  did  he  buy  of  each  kind  ? 

Ans.  14  calves,  42  sheep,  and  98  lambs. 

3.  A  man  paid  $165  to  55  laborers,  consisting  of  men,  women, 
and  boys;  to  the  men  he  paid  $5  a  week,  to  the  women  $1  a  week, 
nnd  to  the  boys  $  £  a  week  ;  how  many  were  there  of  each  ? 

Ans.  30  men,  5  women,  and  20  boys. 


INVOLUTION.  379 


INVOLUTION. 

A  Power  is  the  product  arising  from  multiplying  a 
number  by  itself,  or  repeating  it  any  number  of  times  as  a  factor," 

64O«  Involution  of  the  process  of  raising  a  number  to  a  given 
power. 

®-U.   The  Square  of  a  number  is  its  second  power. 

G42;  The  Cube  of  a  number  is  its  third  power. 

643.   In  the  process  of  involution,  we  observe, 

L  That  the  exponent  of  any  power  is  equal  to  the  number  of 
times  the  root  has  been  taken  as  a  factor  in  continued  multiplica- 
tion. Hence 

II.  The  product  of  any  two  or  more  powers  of  the  same  num- 
ber is  the  power  denoted  by  the  sum  of  their  exponents,  and 

III.  If  any  power  of  a  number  be  raised  to  any  given  power, 
the  result  will  be  that  power  of  the  number  denoted  by  the  pro- 
duct of  the  exponents. 

1.  What  is  the  5th  power  of  6  ? 

OPERATION.  ANALYSIS.    W& 

6x6x6x6x6  =  7776,  Ans.  multiply  G  by  ifc- 

Or,  G2lf»  and  this  pro- 

6  x  6  =  62  =  36  duct  bv  G>  and  G(> 

3G  x  (5  _  53  __  216  on»  until    6    hag 

63  x  6?  =  65  =  216  x  36  =  7776,  Ans.        been  taken  5  times 

in  continued  mul- 

tiplication ;  the  final  product,  7776,  is  the  power  required,  (I).  Or, 
we  may  first  form  the  2d  and  3d  powers  ;  then  the  product  of  these 
two  powers  will  be  the  5th  power  required,  (II). 

2-  What  is  the  6th  power  of  12  ? 

OPERATION. 

ANALYSIS.     We  find  the  cube  of 

the  second  power,  which  must  bo 
144>  =  2980984,  Ans. 


644,   Hence  for  the  involution  of  numbers  we  have  the  fol- 
lowing 


380  INVOLUTION. 

RULE.  L  Multiply  the  given  number  by  itself  in  continued 
multiplication,  till  it  has  been  taken  as  many  times  as  a  factor  as 
there  are  units  in  the  exponent  of  the  required  power.  Or, 

II,  Multiply  together  two  or  more  powers  of  the  given  number, 
the  sum  of  whose  exponents  is  equal  to  the  exponent  of  the  required 
power.  Or, 

III  Raise  some  power  of  the  given  number  to  such  a  power 
^that  the  product  of  the  two  exponents  shall  be  equal  to  the  exponent 
of  the  required  power. 

NOTES.  —  1.  A  fraction  is  involved  to  any  power  by  involving  each  of  its 
terms  separately  to  the  required  power. 

2    Mixed  numbers  should  be  reduced  to  improper  fractions  before  involution. 

3.  When  the  number  to  be  involved  is  a  decimal,  contracted  multiplication 
may  be  applied  with  great  advantage. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  square  of  79?  Ans    6241. 

2.  What  is  the  cube  of  25.4  ?  Ans.  16387.064. 

3.  What  is  the  square  of  1450  ? 

4.  Raise  16  J  to  the  4th  power.  Ans.  79659f  §1 

5.  Raise  2  to  the  20th  power.  Ans.  1048576 

6.  Raise  .4378565  to  the  8th  power,  reserving  5  decimals 

Ans.  .00135  -b 

7.  Raise  1.052578  to  the  6th  power,  reserving  4  decimals 

Ans.  1.3600  ± 

8.  Involve  .029  to  the  5th  power  ? 

Ans.  .000000020511149. 
Find  the  value  of  each  of  the  following  expressions : 

9.  4.367*  Ans.  363.691178934721, 
10    (J)3.  Ans    §13. 
11.   (2J)»                                                                   Ans.   »7«ft. 
12    4.63  x  253                                                   Ans.  1520875. 
13.  (6f  )4  — 7.25a.                                       14    (8£)3  x  2.52 

15.  I  of  (j)3  of  (34)'.  Ans.  5f . 

NOTE.—  Cancel  like  powers  of  the  same  factor. 

16.  7«-±_3.08. 

17    (4"  x  56  x  12s)  -f-  (4*  x  10*  x  32).  Ans.  1200 


EVOLUTION.  381 

EVOLUTION. 

645.  A  Root  is  a  factor  repeated  to  produce  a  power ;  thus, 
in  the  expression  7x7x7  =  343,  7  is  the  root  from  which  the 
power,  343,  is  produced. 

646.  Evolution  is  the  process  of  extracting  the  root  of  a 
number  considered  as  a  power ;  it  is  the  reverse  of  Involution. 

Any  number  whatever  may  be  considered  a  power  whose  root 
is  to  be  extracted. 

647.  A  Rational  Root  is  a  root  that  can  be  exactly  obtained. 

648.  A  Surd  is  an  indicated  root  that  can  not  be  exactly  ob- 
tained. 

649.  The  Radical  Sign  is  the  character,  ^/,  which,  placed 
before  a  number,  indicates  that  its  root  is  to  be  extracted. 

6oO.  The  Index  of  the  root  is  the  figure  placed  above  the 
radical  sign,  to  denote  what  root  is  to  be  taken.  When  no  index 
is  written,  the  index,  2,  is  always  understood. 

6«51*  The  names  of  roots  are  derived  from  the  corresponding 
powers,  and  are  denoted  by  the  indices  of  the  radical  sign.  Thus, 
V'lUO  denotes  the  square  root  of  100,  v/100  denotes  the  cube 
root  of  100;  VlUU  denotes  the  fourth  root  of  100;  etc. 

6d2.  Evolution  is  sometimes  denoted  by  a  fractional  exponent, 
the  name  of  the  root  to  be  extracted  being  indicated  by  the  deno- 
minator. Thus,  the  square  root  of  10  may  be  written  10  ;  the 
cube  root  of  10,  10  ,  etc. 

633.  Fractional  exponents  are  also  used  to  denote  both  invo- 
lution arid  evolution  in  the  same  expression,  the  numerator  indi- 
cating the  power  to  which  the  given  number  is  to  be  raised,  and 

the  denominator  the  root  of  the  power  which  is  to  be  taken ;  thus, 

.2. 
7    denotes  the  cube  root  of  the  second  power  of  7,  and  is  the 

same  as  v/72;  so  also  7~  =  v/75. 

634.  In  extracting  any  root  of  a  number,  any  figure  cr  figures 
may  be  regarded  as  tens  of  the  next  inferior  order.     Thus,  in 
2546,  the  2  may  be  considered  as  tens  of  the  3d  order,  the  25  as 
tens  of  the  second  order,  or  the  254  as  tens  of  the  first  order. 


382  EVOLUTION. 

SQUARE   ROOT. 

655.  The  Square  Root  of  a  number  is  one  of  the  two  equal 
factors  that  produce  the  number.     Thus,  the  square  root  of  64  is 
8,  for  8x8==  64. 

To  derive  the  method  of  extracting  the  square  root  of  a  num- 
ber, it  is  necessary  to  determine 

1st.  The  relative  number  of  places  in  a  number  and  its  square  root 

2d.  The  relations  of  the  figures  of  the  root  to  the  periods  of 
the  number. 

3d.  The  law  by  which  the  parts  of  a  number  are  combined  in 
the  formation  of  its  square ;  and 

4th.  The  factors  of  the  combinations. 

656.  The  relative  number  of  places  in  a  given  number  and 
its  square  root  is  shrwn  in  the  following  illustrations. 

Koots  Squares.  Ecots.  Squares. 

1  1  1  1 

9  81  10  1,00 

99  98,01  100  1,00.00 

999          99,80,01  1000      1,00,00,00 

From  these  examples  we  perceive 

1st.  That  a  root  consisting  of  1  place  may  have  1  or  2  places  in  the 
square. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root  adds  2 
places  to  the  square.  Hence, 

I.  If  we  point  of  a  number  into  two-figure  periods,  commencing 
at  the  right  hand,  the  number  of  fall  periods  and  the  left  hand 
full  or  partial  period  will  indicate  the  number  of  places  in  the 
square  root. 

To  ascertain  the  relations  of  the  several  figures  of  the  root  to  the 
periods  of  the  number,  observe  that  if  any  number,  as  2345,  be  de- 
composed at  pleasure,  the  squares  of  the  left  hand  parts  will  be  re 
lated  in  local  value  as  follow3 : 

20002  =   4  00  00  00 

23002  =   5  29  00  00 

23402  =  S  47  56  00 

23452  =    5  49  90  25:  Hence, 

II.  The  square  of  the  first  figure  of  the  root  is  contained  wholly 
in  the  first  period  of  the  power ;  the  square  of  the  first  two  figures 


SQUARE  ROOT.  383 

of  the  root  is  contained  wholly  in  the  first  two  periods  of  the  power  ; 
and  so  on. 

NOTE. -The  periods  and  figures  of  the  root  are  counted  from  the  left  hand. 

The  combinations  in  the  formation  of  a  square  may  be  shown  as 
follows : 

If  wo  take  any  number  consisting  of  two  figures,  as  43,  and  decom- 
pose it  into  two  parts,  40  +  3,  then  the  square  of  the  number  may 
be  formed  by  multiplying  both  parts  by  each  cart  separately :  thus, 

40  +  3 
40  +  3 

120  +  9 
160C  +  120 

43*  =  1600  +  240  +  9  =  1849. 

Of  these  combinations,  we  observe  that  the  first,  1600,  is  the  square 
of  40  the  second,  240,  is  twice  40  multiplied  by  3  ;  and  the  third,  9, 
is  the  square  of  3.  Hence, 

III.  The  square  of  a  number  composed  of  tens  and  units  is 
equal  to  the  square  of  the  tens,  plus  twice  the  tens  multiplied  by  the 
units,  plus  the  square  of  the  units. 

By  observing  the  manner  in  which  the  square  is  formed,  we  per- 
ceive that  the  unit  figure  must  always  be  contained  as  a  factor  in 
both  the  second  and  third  parts ;  these  parts  taken  together,  may 
therefore  be  factored,  thus,  240  +  9  =  (80  +  3)  X  3.  Hence, 

IV.  If  the  square  of  the  tens  be  subtracted  from  the  entire 
square,  the  remainder  will  be  equal  to  twice  the  tens  plus  the  units 
multiplied  by  the  units. 

1.  What  is  the  square  root  of  5405778576  ? 

OPERATION.  ANALYSIS.     Pointing  off  the 

5405778576  (  73524         given  number  into  periods  of 

49  two  figures  each,  the  5  periods 

143              505  show  that  there  will  be  5  fig- 

429  ures  in  the  root,  (I).     Since 

l£Q5~~          7(377  the  square  of  the  first  figure 

7325  of  the  root  is  always  contained 

14709 35^85  wholly  in  the  first  period  of 

29404  the  power,  (II),  we  seek  for  the 

147044"  ~      '~588176  «reatest  sc*ua];e  in  thc  first  ^ 

588176  riod,   54,  which  we   find  by 

trial  to  be  49,  and  we  place 


384  EVOLUTION. 

its  root,  7,  as  the  first  figure  of  the  required  root,  and  regard  it  as 
tens  of  the  next  inferior  order,  (II).  We  now  subtract  49,  the 
square  of  the  first  figure  of  the  root,  from  the  first  period,  54,  and 
bringing  down  the  next  period,  obtain  505  for  a  remainder.  And 
since  the  square  of  the  first  two  figures  of  the  root  is  contained  wholly 
in  the  first  two  periods  of  the  power,  (II),  the  remainder,  505,  must 
contain  at  least  twice  the  first  figure  (tens)  plus  the  second  figure 
(units),  multiplied  by  the  second  figure,  (TV).  New  if  we  could  divide 
this  remainder  by  twice  the  first  figure  plus  the  second,  which  is  one 
of  the  factors,  the  quotient  would  be  the  second  figure,  or  the  other 
factor.  But  since  we  have  not  yet  obtained  the  second  figure,  the 
complete  divisor  can  not  now  be  employed ,  and  we  therefore  write 
twice  the  first  figure,  or  14,  at  the  left  of  505  for  a  trial  divisor,  re- 
garding it  as  tens.  Dividing  the  dividend,  exclusive  of  the  right 
hand  figure,  by  14,  we  obtain  3  for  the  second,  or  trial  figure  of  the 
root,  which  we  annex  to  the  trial  divisor,  14,  making  143,  the  com- 
plete divisor.  Multiplying  the  complete  divisor  by  the  trial  figure 
3,  and  subtracting  the  product  from  the  dividend,  we  have  76  for  a 
remainder.  We  have  now  taken  the  square  of  the  first  two  figures  of 
the  root  from  the  first  two  periods  ;  and  since  the  square  of  the  first 
three  figures  ot  the  root  is  contained  wholly  in  the  first  three  periods, 
(II)  we  bring  down  the  tnird  period,  77.  to  the  remainder.  76,  and 
obtain  for  a  new  dividend  7677,  which  must  contain  at  least  twiceihe 
two  figures  already  found  plus  the  third,  multiplied  by  the  third,  (IV). 
Therefore  to  obtain  the  third  figure,  we  must  take  for  a  new  trial 
divisor  twice  the  two  figures,  73,  considered  as  tens  of  the  next  infe- 
rior order,  which  we  obtain  in  the  operation  by  doubling  the  last  fig- 
ure of  the  last  complete  divisor,  143,  making  146.  Dividing,  we  ob- 
tain 5  for  the  next  figure  of  the  root ;  then  regarding  735  as  tens  of 
the  next  inferior  order,  we  proceed  as  ;n  the  former  steps,  and  thus 
continue  till  the  entire  root,  73524,  is  obtained. 

G57.  From  these  principles  and  illustrations  we  derive  the 
following 

RULE.  1.  Point  off  the  given  number  into  periods  of  two  figures 
each,  counting  from  units' place  toward  the  left  and  right. 

II.  Find  the  greatest  square  number  in  the  left  hand  period,  and 
write  its  root  for  the  first  figure  in  the  root ;  subtract  the  square 
number  from  the  left  hand  period,  and  to  the  remainder  briny 
down  the  next  period  for  a  dividend* 


SQUARE  ROOT.  3S5 

III.  At  the  left  of  the  dividend  write  twice  the  first  figure  of  the 
root ,  for  a  trial  divisor;  d<oide  the  dividend,  exclusive  of  its  right 
hand  figure,  by  the  trial  divisor,  and  write  the  quotient  for  a  trial 
Jig nre  in  the  root. 

I\  .  Annex  the  tnnl  figure  of  the  root  to  the  trial  divisor  for  a 
complete  divisor;  multiply  the  complete  divisor  by  the  trial  figure 
in  the  root,  subtract  the  product  from  the  dividend,  and  to  the 
remainder  bring  down  the  next  period  for  a  new  dividend. 

V.  Multiply  (he  last  figure  of  the  last  complete  divisor  by  2 
add  the  product  to  10  tiuu'S  the  previous  divisor,  for  a  new 
divisor,  with  which  proceed  as  before. 

NOTES. — 1.  If  at  any  time  the  product  be  greater  than  the  dividend,  diminish 
the  trial  figure  of  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  a  cipher  to  the  tria1  divisor,  and  another 
period  to  the  dividend,  and  proceed  as  before. 

3.  If  there  is  a  remainder  after  all   the   periods   have   been   brought  down, 
annex  periods  oi  ciphers,  and  continue  the  root  to  as  many  decimal  places  as 
are  required. 

4.  The  decimal  points  in  the  woik  may  be  omitted,  care  being  taken  to  point 
off  in  the  root  according  to  the  number  of  decimal  periods  used. 

5.  The  square  root  of  a  common   fraction   may  be  obtained  by  extracting  the 
snuare  roots  of  the  numerator  and  denominator  separately,  provided  the  terms 
are  perfect  squares;  otherwise,  the  fraction  may  first  be  reduced  to  a  decimal. 

6.  Mixed  numbers  may  be  reduced  to  the  decimal  form  before  extracting  fhe 
root ;  or,  if  the  denominator  of  the  fraction  is  a  perfect  square,  to  an  improper 
fraction. 

7.  The  pupil  will  acquire   greater  facility,  and  secure  greater  accuracy,  by 
keeping  units  of  like  order  under  each   other,  and   each  divisor  opposite  the 
corresponding  dividend,  as  shown  in  the  operation. 

EXAMPLES   FOR   PRACTICE. 

1.  What  is  the  square  root  of  315844  ?  Ans.  562. 

2.  What  is  the  square  root  of  152399025  ?       Ans.  12345. 

3.  What  is  the  square  root  of  56280004  ?     Of  597  ? 

4.  What  is  the  square  root  of  10795.21  ?  Ans.  103.9. 

5.  What  is  the  square  root  of  58.14061  ?  Ans.  7.62£. 
Find  the  values  of  the  following  expressions : 

6.  %/.OUOU31G969.  Ans.  .00563. 

7.  >/3858.07694409~64.  Ans.  62.11342. 

8.  %/£  Ans.  .745355  +  . 

9.  x/99225  —  63504.  10.  v/,126736—  >/.045369. 

11-  ^»f  X  V^fg.  Ans.  jf. 

12.  %/81a  x  625»  x  2*.  Ana.  202500. 

33          z 


386  EVOLUTION. 

CONTRACTED    METHOD. 

G«58.    1.  Find  the   square   root   of  8,  correct  to  6  decimal 
places. 

OPERATION.  ANALYSIS.     Extracting  the  square 

|2.828427-f-,  Ans.     root  in  the  usual  way  until  we  have 

8  000000  obtained   the   4  places,  2.828,    the 

4  corresponding  remainder  is  2416,  and 

48          40Q  the  next  trial  divisor,  with  the  cipher 

384  omitted,  is  5656.     We  now  omit  to 

562          1600  bring  down  a  period  of  ciphers  to 

1124  the  remainder,  thus  contracting  the 

5~648 47600  dividend  2  places ;  and  we  contract 

45184  the  divisor  an  equal  number  of  places 

5656  2416*  ^y  omitting  to  annex  the  trial  figure 

2262  °f  tne  root»  an(*  regarding  the  right 

"566  hand  figure,  6,  as  a  rejected  or  re- 

113  dundant  figure      We  now  divide  as 

cZ  77  in  contracted  division  of  decimals, 

4Q  (226),  bringing  down  each  divisor  in 

its  place,  with  one  redundant  figure 

increased  by  1  when  the  rejected  figure  is  5  or  more,  and  carrying  the 
tens  from  the  redundant  figure  in  multiplication.  We  observe  that 
the  entire  root,  2.828427+5  contains  as  many  places  as  there  are  places 
in  the  periods  used.  Hence  the  following 

RULE.  I  If  necessary,  annex  periods  of  ciphers  to  the  given 
number,  and  assume  as  many  figures  as  there  are  places  required 
in  the  root;  then  proceed  in  the  usual  manner  until  all  the  assumed 
figures  have  been  employed ',  omitting  the  remaining  figures,  if  any. 
II.  Form  the  next  trial  divisor  as  usual,  but  omit  to  annex  to  it 
the  trial  figure  of  the  root,  reject  one  figure  from  the  right  to  form 
each  subsequent  divisor,  and  in  multiplying  regard  the  right  hand 
figure  of  each  contracted  divisor  as  redundant. 

NOTES. — 1.  If  the  rejected  figure  is  5  or  more,  increase  the  next  left  hand 
figure  by  1. 

2.  Always  take  full  periods,  both  of  decimals  and  integers. 

EXAMPLES    FOR   PRACTICE. 

1,  Find  the  square  root  of  32  correct  to  the  seventh  decimal 
place.  Ans.  5.6568542+ . 


CUBE  ROOT.  387 

x  2.  Find  the  square  root  of  12  correct  to  the  seventh  decimal 
place-  Ana.  3.4641016+ . 

3.  Find  the  square  root  of  3286.9835  correct  to  the  fourth 
decimal  place.  Ans.  57.3322  +  . 

4  Find  the   square   root  of  .5  correct  to   the   sixth  decimal* 
place.  Ans.  .745355  +  . 

5  Find  the   square   root  of  6j  correct  to  the   sixth  decimal 
place.  Ans.  2.563479  +  . 

6.  Find  the  square  root  of  1.068  correct  to  the  sixth  decimal 
place.  Ans.   1.156817  +  . 

7.  Find  the  value  of  1.01253   correct  to  the  fourth  decimal 
place.  Ans.    1.0188+. 

8.  Find  the  value  of  1.023375*  correct  to  the  sixth  decimal 
place.  Ans.  1.011620  ±. 

CUBE  ROOT. 

G5O.  The  Cube  Root  of  a  number  is  one  of  the  three  equal 
factors  that  produce  the  number.  Thus,  the  cube  root  of  343  is 
7,  since  7x7x7  =  343. 

To  derive  the  method  of  extracting  the  cube  root  of  a  number, 
it  is  necessary  to  determine 

1st.  The  relative  number  of  places  in  a  given  number  and  its 
cube  root. 

2d.  The  relations  of  the  figures  of  the  root  to  the  periods  of 
the  number. 

3d.  The  law  by  which  the  parts  of  a  number  are  combined  in 
the  formation  of  a  cube  j  and 

4th.  The  factors  of  these  combinations. 

OOO.  The  relative  number  of  places  in  a  given  -number  and 
its  cube,  is  shown  in  the  following  illustrations : 

Roots.  Cubes.  Roots.  Cubes. 

11  11 

9  729  10  1,000 

99  907,299  100  1,000,000 

999        997,002,999  1000        1,000,000,000 

From  these  examples,  we  perceive, 


8S8  EVOLUTION. 

1st.  That  a  root  consisting  of  1  place  may  have  from  1  to  3  places 
in  the  cube. 

2d.  That  in  all  cases  the  addition  of  1  place  to  the  root  adds  3 
places  to  the  cube.  Hence, 

I.  If  we  point  off  a  number  into  three-figure  periods,  com- 
mencing at  the  right  hand,  the  number  of  full  periods  and  the  left 
hand  full  or  partial  period  will  indicate  the  number  of  places  in 
the  cube  root. 

To  ascertain  the  relations  of  the  several  figures  of  the  root  to  the 
periods  oi  the  number,  observe  that  if  any  number,  as  5423,  be  de- 
composed, the  cubes  of  the  parts  will  be  related  in  local  value,  as 
follows : 

5000s  =  125  000  000  000 
5400s  =  157  464  000  000 
5420s  ==  159  220  088  000 
5423s  =  159  484  621  967.  Hence, 

II  The  cube  of  the  first  figure  of  the  root  is  contained  wholly  in 
the  first  period  of  the  power  ;  the  cube  of  the  first  two  figures  of 
the  root  is  contained  wholly  in  the  first  two  periods  of  the  power; 
and  so  on 

To  learn  the  combinations  of  tens  and  units  in  the  formation  of  a 
cube,  take  any  number  consisting  of  two  figures,  as  54,  and  decom- 
pose it  into  two  parts,  50+4;  then  having  formed  the  square  by  656, 
III,  multiply  each  part  of  this  square  by  the  units  and  tens  of  54 
separately,  thus, 

542  =  502  +  2  x  50  x  4  +  4» 

50+4 

502  x  4  +  2  x  50  x  42  +  43 
503  +  2  X  50*  x  4  +         50  x  4* 

543  =  503  +  3  X  502  X  4  +  3  x  50  X  42  +  43  =  156924 

Of  these  combinations,  the  first  is  the  cube  of  50,  the  second  is  3 
times  the  square  of  50  multiplied  by  4,  the  third  is  3  times  50  multi- 
plied by  the  square  of  4,  and  the  fourth  is  the  cube  of  4.  Hence, 

III.  The  cube  of  a  number  composed  of  tens  and  units  is  equal 
to  the  cube  of  the  tens,  plus  three  times  the  square  of  the  tens  multi- 
plied by  the  units,,  plus  three  times  the  tens  multiplied  by  the  square 
of  the  units,  plus  the  cube  of  the  units. 

By  observing  the  manner  in  which  the  cube  is  formed,  we  perceive 
that  each  of  the  last  three  parts  contains  the  units  as  a  factor;  these 


CUBE  ROOT.  389 

parts,  considered  as  one  number,  may  therefore  be  separated  into  two 
factors,  thus, 

(3  x  502  +  3  X  50  X  4  +  42)  X  4.     Hence, 

IV.  //'  the  cube  of  the  tens  be  subtracted  from  the  entire  cube, 
the  remainder  u'ill  be  composed  of  two  factors,  one  of  which  will  be 
three  times  the  square  of  the  tens  plus  three  times  the  tens  multiplied 
by  the  units  plus  the  square  of  the  units  ;  and  the  other }  the  units. 

1.  What  is  the  cube  root  of  145780726447  ? 

OPERATION. 

145780726447(5263,  Ans. 


I 

II 

125 

152 

304 

7500        20780 
7804        15608 

1566 

9396 

811200      5172726 
Si.0596      4923576 

1  i~t~OO 

—  •>  \c\ 

83002800    249150- 

v  •••>  i  \  ~  i  \  i  in      o  mi  z.1  \ 

15783  47349  cS3050l49  249150447 
ANALYSIS.  Pointing  off  the  given  number  into  periods  of  3  figures 
each,  the  four  periods  show  that  there  will  be  four  figures  in  the  root, 
(I).  Since  the  cube  of  the  first  figure  of  the  root  is  contained  wholly 
in  the  first  period  of  the  power,  (II),  we  seek  the  greatest  cube  in  the 
first  period,  145,  which  we  find  by  trial  to  be  125,  and  we  place  its 
root,  5,  for  the  first  figure  of  the  required  root,  and  regard  it  as  tens 
of  the  next  inferior  order,  (654).  We  now  subtract  125,  the  cube 
of  this  figure,  from  the  first  period,  145,  and  bringing  down  the  next 
period,  obtain  20780  for  a  dividend.  And  since  the  cube  of  the  first 
two  figures  of  the  root  is  contained  wholly  in  the  first  two  periods 
of  the  power,  (II),  the  dividend,  20780,  must  contain  at  least  the 
product  of  the  two  factors,  one  of  which  is  three  times  the  square  of 
the  first  figure  (tens),  plus  three  times  the  first  figure  multiplied  by 
the  second  (units),  plus  the  square  of  the  second  ;  and  the  other,  the 
second  fiyure  (IV).  Now  if  we  could  divide  this  dividend  by  the  first 
of  these  factors,  the  quotient  would  be  the  other  factor,  or  the  seconcl 
figure  of  the  root.  But  as  the  first  factor  is  composed  in  part  of  the 
second  figure,  which  we  have  not  yet  found,  we  can  not  now  obtain  the 
complete  divisor ;  and  we  therefore  write  three  times  the  square  of 
the  first  figure,  regarded  as  tens,  or  502  X  3  —  7500,  at  the  left  of  the 
dividend,  for  a  trial  divisor.  Dividing  the  dividend  by  the  trial 
divisor,  we  obtain  2  for  the  second,  or  trial  figure  of  the  root.  To 
33* 


390  EVOLUTION. 

complete  the  divisor,  we  must  add  to  the  trial  divisor,  as  a  correction, 
three  times  the  tens  of  the  root  already  found  multiplied  by  the  units, 
plus  the  square  of  the  units,  (IV).  But  as  50  X  3  X  £  -f  22  ==  (50  X 
3  4-  2)  X  2,  we  annex  the  second  figure,  2,  to  three  times  the  first 
figure,  5,  and  thus  obtain  50  X  3  +  2  =  152,  the  first  factor  of  the 
correction,  which  we  write  in  the  column  marked  I.  Multiplying 
this  result  by  the  2,  we  have  304,  the  correction,  which  we  write  in 
the  column  marked  II.  Adding  the  correction  to  the  trial  divisor,  we 
obtain  7804.  the  complete  divisor.  Multiplying  the  complete  divisor 
by  the  trial  figure  of  the  root,  subtracting  the  product  from  the 
dividend,  and  bringing  down  the  next  period,  we  have  5172726  for 
a  dividend. 

We  have  now  taken  the  cube  of  the  first  two  figures  of  the  root 
considered  as  tens  of  the  next  inferior  order,  from  the  first  three 
periods  of  the  number  ;  and  since  the  cube  of  the  first  three  figures 
of  the  root  is  contained  wholly  in  the  first  three  periods  of  the  power, 
(II),  the  dividend,  5172726  must  contain  at  least  the  product  of  the 
two  factors,  one  of  which  is  three  times  the  square  of  the  first  two 
figures  of  the  root  (regarded  as  tens  of  the  next  order)  plus  three 
times  the  first  two  figures  multiplied  by  the  third,  plus  the  square  of  the 
third;  and  the  other,  the  third  figure,  (IV).  Therefore,  to  obtain  the 
third  figure,  we  must  use  for  a  trial  divisor  three  times  the  square  of 
the  first  two  figures,  52,  considered  as  tens.  And  we  observe  that  the 
significant  part  of  this  new  trial  divisor  may  be  obtained  by  adding 
the  last  complete  divisor,  the  last  correction,  and  the  square  of  the 
last  figure  of  the  root,  thus  : 

7804  =  (502  x  3)  +    (50  x  3  X  2)  -f  22 

304  =  50  x  3  X  2  -f  22 

4=  & 


8112  =  (502          -f     00  x  2  +  2*)x3  =  522x  3 

This  number  is  obtained  in  the  operation  without  re-writing  the 
partd,  by  adding  the  square  of  the  second  root  figure  mentally,  and 
combining  units  of  like  order,  thus  :  4,  4,  and  4  are  12,  and  we  write 
the  unit  figure,  2,  in  the  new  trial  divisor  ;  then  1  to  carry  and  0 
is  1  ;  then  3  and  8  are  11,  etc.  Annexing  two  ciphers  to  the  8112, 
because  52  is  regarded  as  tens  of  the  next  order,  and  dividing  by  this 
new  trial  divisor,  811200,  we  obtain  6,  the  third  figure  in  the  root. 
To  complete  the  second  trial  divisor,  after  the  manner  of  completing 
the  first,  we  should  annex  the  third  figure  of  the  root,  6,  to  three 
times  the  former  figures,  52,  for  the  first  factor  of  the  correction. 


CUBE  ROOT.  391 

But  as  we  have  in  column  I  three  times  5  with  the  2  annexed,  or  152, 
we  need  only  multiply  the  last  figure,  2,  by  3,  and  annex  the  third 
figure  of  the  root,  6,  which  gives  1566,  the  first  factor  of  the  correc- 
tion sought,  or  the  second  term  in  column  I.  Multiplying  this  number 
by  the  6,  we  obtain  9396,  the  correction  sought ;  adding  the  correction 
to  the  trial  divisor,  we  have  820596,  the  complete  divisor  ;  multiplying 
the  complete  divisor  by  the  6,  subtracting  the  product  from  the  divi- 
dend, and  bringing  down  the  next  period,  we  have  249150447  for  a 
new  dividend  We  may  now  regard  the  first  three  figures  of  the  root, 
526,  as  tens  of  the  next  inferior  order,  and  proceed  as  before  till  the 
entire  root,  5263,  is  extracted. 

OO1.    From  these  principles  and  illustrations  we  deduce  the 

following 

RULE.     I.  Point   off  the  given    number  into  periods  of  three 
figures  each,  counting  from  units" place  toward  the  left  and  fight. 

II.  Find  the  greatest  cube  that  does  not  exceed  the  left  hand 
period ,  and  write  its  root  for  the  first  figure  in  the  required  root; 
subtract  the  cube  from  the  left  hand  period,  and  to  the  remainder 
briny  down  the  next  period  for  a  dividend. 

III.  At  the  left  of  the  dividend  write  three  times  the  square  of 
the  first  figure  of  the  root,  and  annex  two  ciphers,  for  a  trial  di- 
visor ;  divide  the  dividend  by  the  trial  divisor,  and  write  the  quo- 
tient for  a  trial  figure  in  the  root. 

IV.  Annex  the  trial  figure  to  three  times  the  former  figure,  and 
write  the  result  in  a  column  marked  I,  one  line  below  the  trial 
divisor,   multiply  this  term    by   the    trial  figure,  and  write  the 
product  on  the  same  line  in  a  column  marked  II;  add  this  term 
as  a  correction  to  the  trial  divisor,  and  the  result  will  be  the  com- 
plete  divisor. 

V.  Multiply  the  complete  divisor  by  the  trial  figure ;  subtract 
the  product  from  the  dividend,  and  to  the  remainder  bring  down 
the  next  period  for  a  new  dividend. 

VI.  Add  the  square  of  the  last  figure  of  the  root>  the  last  term 
in  column  II,  and  'he  complete  divisor  together,  and  annex  two 
ciphers,  for  a  new  trial  divisor}  with  which  obtain  another  trial 
figure  in  the  root. 


392  EVOLUTION. 

VII.  Multiply  the  unit  figure  of  the  last  term  in  column  I  by 
3,  and  annex  the  trial  fijure  of  the  root  for  the  next  term  of 
column  I;  multiply  this  result  by  the  trial  fiyure  *of  the  root  for 
the  next  term  of  column  II;  add  this  term  to  the  trial  divisor  for 
a  complete  divisor,  with  which  proceed  as  before. 

NOTES. — 1.  If  at  any  time  the  product  be  greater  than  the  dividend,  diminish 
the  trial  figure  <.f  the  root,  and  correct  the  erroneous  work. 

2.  If  a  cipher  occur  in  the  root,  annex  two  more  ciphers  to  the  trial  divisor, 
and  another  period  to  the  dividend;  then  proceed  as  before  with  column  I,  an, 
nexmg  both  cipher  and  trial  figure. 

EXAMPLES   FOR   PRACTICE 

1.  "What  is  the  cube  root  of  389017  ?  Am.  73. 

2.  What  is  the  cube  root  of  44361864  ?  Ans.  354. 

3.  What  is  the  cube  root  of  10460353203?  Ans.  2187. 
4.*  What  is  the  cube  root  of  98867482624  ?  Ans.  4624 

5.  What  is  the  cube  root  of  30.625  ?         Ans.  3  12866  +. 

6.  What  is  the  cube  root  of  111  J  ?  Ans    4.8076  f 

7.  What  is  the  cube  root  of  .000148877?  Ans    .053. 
Find  the  values  of  the  following  expressions. 

8.  Vl22615327232'f  Ans.  4968. 
9-  ^^7134217728"?  Ans.  8. 

10.  V39304*"?  Ans.  1156. 

11-   ^im  X  ^mT?  Ans.  ||. 

12  How  much  does  the  sum  of  the  cube  roots  of  50  and  31 
exceed  the  cube  root  of  their  sum?  Ans.  2.4986  -f . 

CONTRACTED    METHOD. 

0G2.  In  applying  contracted  decimal  division  to  the  extrac- 
tion of  the  cube  root  of  numbers,  we  observe, 

1st.  For  each  new  figure  in  the  root,  the  terms  in  the  operation 
extend  to  the  right  3  places  in  the  column  of  dividends,  2  places 
in  the  column  of  divisors,  and  1  place  in  column  I.  Hence, 

2d.  If  at  any  point  in  the  operation  we  omit  to  bring  down  new 
periods  in  the  dividend,  we  must  shorten  each  succeeding  divisor 
1  place,  and  each  succeeding  term  in  column  I,  2  places. 

1.   What  is  the  cube  root  of  189,  correct  to  8  decimal  places  ? 


CUBE  ROOT. 


393 


ANALYSIS.  We 
proceed  by  the  usual 
method  to  extract 
the  cube  root  of  the 
given  number  until 
we  have  obtained 
the  three  figures, 
5.73  :  the  corres- 
ponding remainder 
is  867483,  and  the 
next  trial  divisor 
with  the  ciphers 
omitted  is  984987. 
We  now  omit  to 
bring  down  a  period 
of  ciphers,  thus  con- 
tracting the  divid- 
end 3  places ;  and 
we  contract  the  di- 
visor an  equal  num- 
ber of  places  by 
emitting  to  annex  the  two  ciphers,  and  regarding  the  right  hand 
figure,  7,  as  a  redundant  figure.  Then  dividing,  we  obtain  8  for  the 
next  figure  of  the  root.  To  complete  the  divisor,  we  obtain  a  correc- 
tion, 1375,  contracted  2  places  by  omitting  to  anrex  the  trial  figure 
of  the  root,  8,  to  the  first  factor,  1719,  and  regarding  the  right  hand 
figure,  9,  as  redundant  in  multiplying.  Adding  the  contraction  to 
the  contracted  divisor,  we  have  the  complete  divisor,  986362,  the  right 
hand  figure  being  redundant.  Multiplying  by  8  and  subtracting  the 
product  from  the  dividend,  we  have  78393  for  a  new  dividend.  Then 
to  form  the  new  trial  divisor,  we  disregard  the  square  of  the  root 
figure,  8.  because  this  square  consists  of  the  same  orders  of  units  as 
the  two  rejected  places  in  the  divisor;  and  we  simply  add  the  cor- 
rection, 1375,  and  the  complete  divisor,  986362,  and  rejecting  1  figure, 
thus  obtain  98774,  of  which  the  right  hand  figure,  4,  is  redundant. 
Dividing,  we  obtain  7  for  the  next  root  figure.  Rejecting  2  places 
from  the  last  term  in  column  I,  we  have  17  for  the  next  contracted 
term  in  this  column.  We  then  obtain,  by  the  manner  shown  in  the 
former  step,  the  correction  12,  the  complete  divisor,  98786,  the  prod- 
uct, 69150,  and  the  new  dividend,  9243.  We  then  obtain  the  new  trial 


I    II 

OPERATION. 

|5.73879355db,  Ans. 

189.000000 
125 

157    1099 

7500   64  000 
8599    60  193 

1713   5139 

974700   3  807000 
979839   2939517 

1719   1375 

984987    867483* 
986362    789090 

17      12 

98774     78393 
98786     69150 

9880       9243 

8892 

988        35  i 
296 

99          55 

50 

10          5 
5 

394  EVOLUTION. 

divisor,  9880;  and  as  column  I  is  terminated  by  rejecting  the  two 
places,  17,  we  continue  the  contracted  division  as  in  square  root,  and 
thus  obtain  the  entire  root,  5.73879355  rb,  which  is  correct  to  the  last 
decimal  place,  and  contains  as  many  places  as  there  are  places  in  the 
periods  used.  Hence  the  following 

RULE.  I.  If  necessary,  annex  ciphers  to  the  given  number,  and 
assume  as  many  figures  as  there  are  places  required  in  the  root ; 
then  proceed  by  the  usual  method  until  all  the  assumed  figures  have 
been  employed. 

II.  Form  the  next  trial  divisor  as  usual,  but  omit  to  annex  the 
two  ciphers,  and  reject  one  place  in  forming  each  subsequent  trial 
divisor. 

Ill  In  completing  the  contracted  divisors,  omit  at  first  to  annex 
the  trial  figure  of  the  root  to  the  term  in  column  I,  and  reject  2 
places  in  forming  each  succeeding  term,  in  this  column. 

IV.  In  multiplying,  regard  the  right  hand  figure  of  each  con- 
tracted term,  in  column  I  and  in  the  column  of  divisors,  as  redund- 
ant. 

NOTES. — 1.  After  the  contraction  commences,  the  square  of  the  last  root  figure 
is  disregarded  in  forming  the  new  trial  divisors. 
2.  Employ  only  full  periods  in  the  number. 

EXAMPLES    FOR   PRACTICE. 

1.  Find  the  cube  root  of  24,  correct  to  7  decimal  places. 

Ans.  2.8844992  db. 

2.  Find  the  cube  root  of  12000.812161,  correct  to  9  decimal 
places.  Ans.  22.89480 1#*4  dr. 

3.  Find  the  cube  root  of  .171467,  correct  to  9  decimal  places. 

Ans.  .555554730  rt. 

4.  Find  the  cube  root  of  2. 42999  correct  to  5  decimal  places. 

Ans.  1.34442±. 

5.  Find  the  cube  root  of  19.44,  correct  to  4  decimal  places. 

Ans.  2.6888    =b. 

6.  Find  the  value  of  3/f"  to  6  places.          Ans.  .941035  ±. 

7.  Find  the  value  of  ^.571428  to  9  places. 

Ans.  .829826686  ±. 


HOOT'S   OK    ANY    DKftKXK.  395 

8.  Find  the  value  of  N/1.08674325'2  to  7  places. 

Am.  1.057023  db. 

9.  Find  the  value  of  1.05s  to  7  places. 

Ant.  1.084715  ±. 

ROOTS   OF   ANY   DEGREE. 

663.  Any  root  whatever  may  be  extracted  by  means  of  the 
square  and  cube  roots,  as  will  be  seen  in  the  two  cases  which  follow. 

CASE    I. 

664.  When  the  index  of  the  required  root  contains 
no  other  factor  than  2  or  3. 

We  have  seen  that  if  we  raise  any  power  of  a  given  number  to 
any  required  power,  the  result  will  be  that  power  of  the  given 
number  denoted  by  the  product  of  the  two  indices,  (643,  III). 
Conversely,  if  we  extract  successively  two  or  more  roots  of  a  given 
number,  the  result  must  be  that  root  of  the  given  number  denoted 
by  the  product  of  the  indices. 

1.  What  is  the  6th  root  of  2176782336  ? 

OPERATION.  ANALYSIS.     The  index  of  the 

6  =  2x3  required  root  is  6  =  2x3;  we 

v/2176782336  =  46657  therefore  extract  the  square  root 

v/ 4665 6  =  36,  Ans.         °f  ^e   given   number,  and   the 

cube  root  of  this  result,  and  oi> 
Or'  tain  36,  which  must  be  the  6th 

^2176782336  =  1296  root  required.     Or,  we  first  find 

N/1296  =  36    Ans.         the  cube  root  of  the  given  num- 

ber, and  then  the  square  root  of 
the  result,  as  in  the  operation.     Hence  the  following 

RULE.  Separate  the  index  of  the  required  root  into  its  prime 
factors,  and  extract  successively  the  roots  indicated  by  the  several 
factors  obtained ;  the  final  result  will  be  the  required  root. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  6th  root  of  6321363049  ?  Ans.  43. 

2.  What  is  the  4th  root  of  5636405776?  Ans.  274. 


396  EVOLUTION. 

3.  What  is  the  8th  root  of  1099511627776?  Ans.  32. 

4.  What  is  the  6th  root  of  25632972850442049  ?    Ans. "  543. 

5.  What  is  the  9th  root  of  1.577635?        Ans.  1 . 051 9.03 +  . 

NOTE. — Extract  the  cube  root  of  the  cube  root  by  the  contracted  method, 
carrying  the  root  in  each  operation  to  6  decimal  places  only. 

6.  What  is  the  12th  root  of  16.3939?  Ans.  1.2(324  +  . 

7.  What  is  the  18th  root  of  104.9617  ?          Ans.  1.2950+. 

CASE   II. 

665.  "When  the  index  of  the  required  root  is  prime, 
or  contains  any  other  factor  than  2  or  3. 

To  extract  any  root  of  a  number  is  to  separate  the  number  into 
as  many  equal  factors  as  there  are  units  in  the  index  of  the  re- 
quired root ;  and  it  will  be  found  that  if  by  any  means  we  can 
separate  a  number  into  factors  nearly  equal  to  each  other,  the 
average  of  these  factors,  or  their  sum  divided  the  number  of  fac- 
tors, will  be  nearly  equal  to  the  root  indicated  by  the  number  of 
factors. 

1.  What  is  the  7th  root  of  308  ? 

OPERATION.  ANALYSIS.     We  first 

£/308  _  2.59+  fincl  by  Case  I,  the  6th 

s/308  =  2  04+  root»  and  a^so  *ke  ^th. 

2.59  +  2.04  =  4.63  r°°t  of  308 ;  and  since 

4.63  -r-       2  =  2.31,  assumed  root.  the  7th  root  must  be 

2.3 16         =  151.93  less  than  the    former 

308  ~  151.93  =  2.0272+  and   greater  than   the 

?k3LVW'°?^15,-8f2         •      *         latter,  we  take  the  ave- 
15.88,2  3^=  2.2696,  1st  approximation.    rage  of  the  tw0j  or  ono 

^l66{fS7=82.253452+  half  of  theirsums,  2.31 

2.2696  x  6  +  2.253452  =  15.871052  and  cal1  li  the  assumed 

15.871052  -r-  7  =  2.267293,  2d  approx. 

the  assumed  root,  2.31, 

to  the  6th  power,  and  divide  the  given  number,  308,  ^y  the  result, 
and  obtain  2.0272+  for  a  quotient ;  we  thus  separate  308  into  7  fac- 
tors, 6  of  which  are  equal  to  2.31,  and  the  other  is  2.0272.  As  these 
7  factors  are  nearly  equal  to  each  other,  the  average  of  them  all  must 
be  a  near  approximation  to  the  7th  root.  Multiplying  the  2.31  by  6, 
adding  the  2.0272  to  the  product,  and  dividing  this  result  by  7,  we 


ROOTS   OF   ANY    DEGREE.  337 

find  the  average  to  be.  2.2696,  which  is  the  first  approximation  to  the 
required  root.  We  next  divide  308  by  the  6th  power  of  2.2696,  and 
obtain  2.253452-}-  for  a  quotient ;  and  we  thus  separate  the  given 
number  into  7  factors,  6  of  which  are  each  equal  to  2.2696,  and  the 
other  is  2.253452.  Finding  the  average  of  these  factors,  as  in  the 
former  steps,  we  have  2.267293,  which  is  the  7th  root  of  the  given 
number,  correct  to  5  decimal  places.  Hence  the  following 

RULE.  I.  Find  by  trial  some  number  nearly  equal  to  the  re- 
quired root,  and  call  tins  the  assumed  root. 

II.  Divide  the  given  number  by  that  power  of  the  assumed  root 
denoted  by  the  index  of  the  required  root  less  1  ;  to  this  quotient 
add  as  many  times  the  assumed  root   as  there  are  units  in  the 
index  of  the  required  root  less  1,  and  divide  the  amount  by  the 
index  of  the  required  root.      The  result  will  be  the  first  approxi- 
mate root  required. 

III.  Take  the  last,  approximation  for  the  assumed   root,  with 
which  proceed  as  with  the  former,  and  thus  continue  till  the  re- 
quired root  is  obtained  to  a  sufficient  degree  of  exactness. 

NOTES. — 1.  The  involution  and  division  in  all  cases  will  be  much  abridged  by 
decimal  contraction. 

2.  If  the  index  cf  the  required  root  contains  the  factor?.  2  or  3,  we  may  first 
extract  the  square  or  cube  root  as  many  times,  successively,  as  these  factors  are 
found  in  the  index,  after  which  we  must  extract  that  root  of  the  result  which  is 
denoted  by  the  remaining  factor  of  the  index.  Thus,  if  the  15th  root  were  re- 
quired, we  should  first  find  the  cube  root,  then  the  5th  root  of  this  result. 

EXAMPLES    FOR    PRACTICE. 

1.  What  is  the  20th  root  of  617  ? 

OPERATIONS 

_20  =   2    x    2    x    5. 
^617  =  24.839485+. 
v'24l*39485  =     4.983923 -K 
v/l9^9T3~  =     1.378L'uti  +  .  Ans. 

2.  What  is  the     5th  root  of  120  ? 

3.  What  is  the     7th  root  of  1.95678  ? 

4.  What  is  the  10th  root  of  743044? 

5.  What  is  the  15th  root  of  15  ? 
6    What  is  the  25th  root  of  100  ? 
7.  What  is  the  5th  root  of  5  ? 

34 


398  SERIES. 

SERIES.* 

666.  A  Series  is  a  succession  of  num  ,ers  so  related  to  each 
other,  that  each  number  in  the  succession  may  be  formed  in  the 
same  manner,  from  one  or  more  preceding  numbers.     Thus,  any 
number  in  the  succession,  2,  5,  8,  11,  14,  is  formed  by  adding  3  to 
the  preceding  number.     Hence,  2,  5,  8,  11,  14,  is  a  series. 

667.  The  Law  of  a  Series  is  the  constant  relation  existing 
between  two  or  more  terms  of  the  series.     Thus,  in  the  series,  3,  7, 
11,  15,  we  observe  that  each  term  after  the  first  is  greater  than  the 
preceding  term  by  4 ;  this  constant  relation  between  the  terms  is 
the  law  of  this  series. 

The  law  of  a  series,  and  the  term  or  terms  on  which  it  depends 
being  given,  any  number  of  terms  of  the  series  can  be  formed. 
Thus,  let  64  be  a  term  of  a  series  whose  law  is,  that  each  term  is 
four  times  the  preceding  term.  The  term  following  64  is  64  x  4, 
the  next  term  64  x  42,  etc.  ;  the  term  preceding  64  is  64  4-  4. 
Hence  the  series,  as  far  as  formed,  is  16,  64,  256,  1024. 

668*  A  series  is  either  Ascending,  or  Descending,  according  as 
each  term  is  greater  or  less  than  the  preceding  term.  Thus,  2,  6, 
10,  14,  is  an  ascending  series;  32,  16,  8,  4,  is  a  descending  series. 

669.  An  Extreme  is  either  the  first  or  last  term  of  a  series. 
Thus,  in  the  series,  4,  7,  10,  13,  the  first  extreme  is  4,  the  last,  13. 

670.  A  Mean  is  any  term  between  the  two   extremes.     Thus, 
in  the  series,  5,  10,  20,  40,  80,  the  means  are  10,  20,  and  40. 

*  The  treatise  of  the  "  METRIC  SYSTEM,"  as  presented  at  some  length  at 
the  close  of  the  editions  of  this  book  published  previous  to  the  year  1875, 
was  of  little  value,  since  the  system  is  scarcely  any  used  in  this  country, 
and  the  symbols  introduced  were  different  from  those  authorized  or  in  use. 

Being  frequently  requested  by  teachers  to  add  or  substitute  an  article 
on  Mensuration,  as  of  much  more  practical  value,  the  editor  has  carefully 
prepared  and  substituted  such  a  treatise  at  the  end  of  this  book,  and 
to  avoid  repetition  and  put  in  more  condensed  form,  has  embodied  in  it 
the  "  applications  of  the  square  and  cube  roots  "  that  intervened  between 
"  Evolution  "  and  "  Series  "  of  former  editions.  So  much  of  the  Metric 
System  as  is  needful  has  also  been  added. 

The  article  on  "  Mensuration  "  is  essentially  the  same  as  that  presented 
in  the  "  Complete  Arithmetic"  of  the  "  Shorter  Course,"  and  it  is  hoped 
will  be  entirely  satisfactory. 


PROGRESSIONS.  399 

671.  An  Arithmetical  or  Equidifferent  Progression  is  a 

series  whose  law  of  formation  is  a  common  difference.  Thus,  in 
the  arithmetical  progression,  3,  7,  11,  15, 19,  each  term  is  formed 
from  the  preceding  by  adding  the  common  difference,  4. 

672.  An  arithmetical  progression  is  an  ascending  or  descend- 
ing series,  according  as  each  term  is  formed  from  the  preceding 
term  by  adding  or  subtracting  the  common  difference.     Thus,  the 
ascending  series,  7,  10,  13,  16,  etc.,  is  an  arithmetical  progression 
in  which  the  common  difference,  3,  is  constantly  added  to  form 
each  succeeding  term;  and  the  descending  series,  20,  17,  14,  11, 
8,  5,  2,  is  an  arithmetical  progression  in  which  the  common  dif- 
ference is  constantly  subtracted-,  to  form  each  succeeding  term. 

673.  A  Geometrical  Progression  is  a  series  whose  law  of 
formation  is  a  common  multiplier.     Thus,  in  the  geometrical  pro- 
gression, 3,  6,  12,  24,  48,  each  term  is  formed  by  multiplying  the 
preceding  term  by  the  common  multiplier,  2. 

674.  A  geometrical  progression  is  an  ascending  or  descending 
series,  according  as  the  common  multiplier  is  a  whole  number  or 
a  fraction.     Thus,  the  ascending  series,  1,  2,  4,  8,  16,  etc.,  is  a 
geometrical   progression   in  which  the  common  multiplier  is  2; 
and  the  descending  series,  32,  16,  8,  4,  2,  1,  £,  £,  etc.,  is  a  geo- 
metrical progression  in  which  the  common  multiplier  is  J. 

675.  The  Ratio  in  a  geometrical  progression  is  the  common 
multiplier. 

676.  In  the  solution  of  problems  in  Arithmetical  or  Geomet- 
rical progression,  five  parts  or  elements  are  concerned,  viz : 

In  Arithmetical  Progression  —        In  Geometrical  Progression  — 

1.  The  first  term  ;  1.  The  first  term ; 

2.  "     last  term ;  2.     "     last  term ; 

3.  "     number  of  terms ;  3.     "     number  of  terms ; 

4.  "     common  difference ;  4.     "     ratio ; 

5.  "     sum  of  the  series.  5.     "     sum  of  the  series. 
The  conditions  of  a  problem  in  progression  may  be  such  as  to 

require  any  one  of  the  five  parts  from  any  three  of  the  four  re- 
maining parts ;  hence,  in  either  Arithmetical  or  Geometrical  Pro- 
gression, there  are  5  X  4  =  20  cases,  or  classes  of  problems,  and 
no  more,  requiring  each  a  different  solution. 


400  SERIES. 

GENERAL  PROBLEMS  IN  ARITHMETICAL  PROGRESSION. 
PROBLEM    I. 

677.  Given,  one  of  the  extremes,  the  common  dif- 
ference, and  the  number  of  terms,  to  find  the  other 
extreme. 

Let  2  be  the  first  term  of  an  arithmetical  progression,  and  3  the 
common  difference ;  then, 

2  =2  =2,  1st  term. 

2+3  =2+  (3  X  1)=   5,  2d      " 

2  +  3  +  3         =2+  (3  X  2)=   8,  3d      " 
2  +  3  +  3  +  3  =  2  +  (3  X  3)  =  11,  4th     " 

From  this  illustration  we  perceive  that,  in  an  arithmetical  pro- 
gression, when  the  series  is  ascending,  the  second  term  is  equal  to  the 
first  term  plus  the  common  difference ;  the  third  term  is  equal  to  the 
first  term  plus  2  times  the  common  difference ;  the  fourth  term  is 
equal  to  the  first  term  plus  3  times  the  common  difference ;  and  so  on. 
In  a  descending  series,  the  second  term  is  equal  to  the  first  term 
minus  the  common  difference ;  the  third  term  is  equal  to  the  first 
minus  2  times  the  common  difference ;  and  so  on.  In  all  cases  the 
difference  between  the  two  extremes  is  equal  to  the  product  of  the 
common  difference  by  the  number  of  terms  less  1.  Hence  the 

RULE.  Multiply  the  common  difference  by  the  number  of  terms 
less  1  •  add  the  product  to  the  given  term  if  it  be  the  less  extreme, 
and  subtract  the  product  from  the  given  term  if  it  be  the  greater 
extreme. 

EXAMPLES   FOR   PRACTICE. 

1.  The  first  term  of  an  arithmetical  progression  is  5,  the  com- 
mon difference  4,  and  the  number  of  terms  8 ;  what  is  the  last 
term  ?  Ans.  33. 

2.  If  the  first  term  of  an  ascending  series  be  2,  and  the  com- 
mon difference  3,  what  is  the  50th  term  ? 

3.  The  first  term  of  a  descending  series  is  100,  the  common 
difference  7,  and  the  number  of  terms  13  ;  what  is  the  last  term  ? 

4.  If  the  first  term  of  an  ascending  series  be  f ,  the  common 
difference  $,  and  the  number,  of  terms  20,  what  is  the  last  term  ? 

Ans.  ?. 


ARITHMETICAL  PROGRESSION.  £01 

PROBLEM    II. 

678.  Given,  the  extremes  and  number  of  terms,  to 
find  the  common  difference. 

Since  the  difference  of  the  extremes  is  always  equal  to  the  common 
difference  multiplied  by  the  number  of  terms  less  1,  (677)>  we  have 
the  following 

RULE.  Divide  the  difference  of  the  extremes  by  the  number  of 
terms  less  1 

EXAMPLES    FOR   PRACTICE. 

1.  If  the  extremes  of  an  arithmetical  series  are  3  and  15,  and 
the  number  of  terms  7,  what  is  the  common  difference? 

Ans.  2. 

2.  The  extremes  are  1  and  51,  and  the  number  of  terms  is  76; 
what  is  the  common  difference  ? 

3.  The  extremes  are  .05  and  .1,  and  the  number  of  terms  is  8; 
what  is  the  common  difference?  Ans.  .00714285. 

4.  If  the  extremes  are  0  and  2i,  and  the  number  of  terms  is 
18;  what  is  the  common  difference  ? 

PROBLEM   III. 

679.  Given,  the  extremes  and  common  difference, 
to  find  the  number  of  terms. 

Since  the  difference  of  the  extremes  is  equal  to  the  common  differ- 
ence multiplied  by  the  number  of  terms  less  1,  (677)>  we  have  the 
following 

RULE.  Divide  the  difference  of  the  extremes  by  the  common 
difference,  and  add  1  to  the  quotient. 

EXAMPLES   FOR   PRACTICE. 

1.  The  extremes  of  an  arithmetical  series  are  5  and  75,  and  the 
common  difference  is  5 ;  what  is  the  number  of  terms  ? 

Am.   15. 

2.  The  extremes  are  |  and  20,  and  the  common  difference   is 
6  £ ;  find  the  number  of  terms. 


402  SERIES. 


3.  The  extremes  are  2.5  and  .25,  and  the  common  difference  is 
.125;  what  is  the  number  of  terms? 

4.  Insert  5  arithmetical  means  between  2  and  37. 


PROBLEM    IV. 

68O.  Given,  the  extremes  and  number  of  terms,  to 
find  the  sum  of  the  series. 

Let  us  take  any  series,  as  2,  5,  8,  11,  14,  and  writing  under  it  the 
same  series  in  an  inverse  order,  add  each  term  of  the  inverted  series 
to  the  term  above  it  in  the  direct  series,  thus : 

2  +    5  +    8  +  11  -f  14  =  40,  once  the  sum, 
14  +  H+    8+    5+    2  =  40,      "      "      " 
16  -f  16  -f  16  +  16  +  16  =  80,  twice  the  sum. 

From  this  we  perceive  that  16,  the  sum  of  the  extremes  of  the  given 
series,  multiplied  by  5,  the  number  of  terms,  equals  80,  which  is  twice 
the  sum  of  the  series  ;  and  80  -r-  2  =  40,  the  sum  of  the  series.  Hence 

RULE.  Multiply  the  sum  of  the  extremes  by  the  number  of 
terms,  and  divide  the  product  by  2. 

EXAMPLES    FOR   PRACTICE. 

1.  Find  the  sum  of  the  series  the  first  term  of  which  is  4,  the 
common  difference  6,  and  the  last  term  40.  Ans.  154. 

2.  The  extremes  are  0  and  250,  and  the  number  of  terms  is 
1000 ;  what  is  the  sum  of  the  series  ? 

3.  A  person  wishes  to  discharge  a  debt  in  11  annual  payments 
such  that  the  last  payment  shall  be  $220,  and  each  payment  greater 
than  the  preceding  by  $17 ;  find  the  amount  of  the  debt,  and  the 
first  payment.  Ans.  First  payment,  $50. 

G81.  By  reversing  some  one  of  the  four  problems  now  given, 
or  by  combining  two  or  more  of  them,  all  of  the. sixteen  remain- 
ing problems  of  Arithmetical  Progression  may  be  solved  or 
analyzed. 


GEOMETRICAL  PROGRESSION.  403 

GENERAL  PROBLEMS  IN  GEOMETRICAL  PROGRESSION, 
PROBLEM    I. 

682.  Given,  one  of  the  extremes,  the  ratio,  and  the 
number  of  terms,  to  find  the  other  extreme. 

Let  3  be  the  first  term  of  a  geometrical  progression,  and  2  the 
ratio :  then, 

3  =3  =    3,  the  1st  term, 

3x2  =3  X2*  =    6,     "  2d      " 

3x2x2         ==  3  X  22  =  12,     "3d      " 
3  X  2  X  2  X  2  =  3  X  23  =  24,     "4th     " 

From  this  illustration  we  perceive  that,  in  a  geometrical  progression, 
the  second  term  is  equal  to  the  first  term  multiplied  by  the  ratio ;  the 
third  term  is  equal  to  the  first  term  multiplied  by  the  second  power 
of  the  ratio;  ths  fourth  term  is  equal  to  the  first  term  multiplied  by 
the  third  power  of  the  ratio ;  and  so  on.  The  same  is  true  whether 
the  ratio  be  an  integer  or  fraction.  Hence  the  following 

RULE.  I.  If  the  given  extreme  be  the  first  term,  multiply  it  by 
that  power  of  the  ratio  indicated  by  the  number  of  terms  less  1 ; 
the  result  will  be  the  last  term. 

II.  If  the  given  extreme  be  the  last  term,  divide  it  by  that  power 
of  the  ratio  indicated  by  the  number  of  terms  less  1;  the  result 
will  be  the  first  term. 

EXAMPLES    FOR   PRACTICE. 

1.  The  first  term  of  a  geometrical  series  is  6,  the  ratio  4,  and 
the  number  of  terms  6 ;  find  the  last  term.  Ans.  6144. 

2.  The  last  term  of  a  geometrical  series  is  192,  the  ratio  2,  and 
the  number  of  terms  7 ;  what  is  the  first  term  ? 

8.  If  the  first  term  be  6,  the  ratio  -J,  and  the  number  of  terms 
8,  what  is  the  last  term  ? 

4.  The  first  term  is  25,  the  ratio  ^,  and  the  number  of  terms 
5 ;  what  is  the  last  term  ?  Ans.  . 


404  SERIES. 

PROBLEM    TI. 

683.  Given,  the  extremes  and  number  of  terms,  to 
find  the  ratio. 

Since  the  last  term  is  always  equal  to  the  first  term  multiplied  by 
that  power  of  the  ratio  indicated  by  the  number  of  terms  less  1, 
(682),  we  have  the  following 

RULE.  Divide  the  last  term  by  the  first ,  and  extract  that  root 
of  the  quotient  indicated  by  the  number  of  terms  less  1  ;  the  result 
iv ill  be  the  ratio. 

EXAMPLES    FOR   PRACTICE. 

1.  The  extremes  are  2  and  512,  and  the  number  of  terms  is  5; 
what  is  the  ratio  ?  Ans.  4. 

2  The  extremes  are  -£•$  and  45T9g,  and  the  number  of  terms  is 
8;  what  is  the  ratio? 

3.  The  extremes  are  7  and  .0112,  and  the  number  of  terms  is 
5 ;  what  is  the  ratio  ?  Ans.  5. 

4.  Insert  3  geometrical  means  between  8  and  5000. 

PROBLEM    III. 

684.  Given,  the  extremes  and  ratio,  to  find  the  num- 
ber of  terms. 

Since  the  quotient  of  the  last  term  divided  by  the  first  term  is 
equal  to  that  power  of  the  ratio  indicated  by  the  number  of  terms 
less  1,  (683),  we  have  the  following 

RULE.  Divide  the  last  term  by  the  first,  divide  this  quotient  by 
the  ratio,  and  the  quotient  thus  obtained  by  the  ratio  again,  and  so 
on  in  successive  division,  till  the  final  quotient  is  I.  The  number 
of  times  the  ratio  is  used  as  a  divisor,  plus  1,  is  the  number  of 
terms. 

EXAMPLES    FOR   PRACTICE. 

1.  The  extremes  are  2  and  1458,  and  the  ratio  is  3;  what  is 
the  number  of  terms  ?  Ans.  7. 

2.  The  first  term  is  .1,  the  last  term  100,  and  the  ratio  10;  find 
the  number  of  terms. 


GEOMETRICAL  PROGRESSION.  405 

3.  The  first  term  is  g^,  the  last  term  ^,  and  the  ratio  2;  what 
is  the  number  of  terms  ? 

4.  The  extremes  are  196608  and  6,  and  the  ratio  is  i ;  what  is 
the  number  of  terms  ?  Ans.  6. 

PROBLEM    IV. 

685.  Given,  the  extremes  and  ratio,  to  find  the  sum 
of  the  series. 

Let  us  take  the  series  5  +  20  +  80  +  320=425,  multiply  each  term 
by  the  ratio  4,  and  from  this  result  subtract  the  given  series  term  from 
term,  thus : 

20  -f  80  +    320  +  1280  =  1700,  four  times  the  series, 

5  4-  20  +  80  +    320  =  425_,_once  the  series, 

1280  —    5     =  1275,  three  times  the  series, 
Then  1275  ~    3     =   425,  once  the  series. 

Hence  the 

RULE.   Multiply  the  greater  extreme,  by  the  ratio,  subtract  the 

less  extreme  from  the  product,  and  divide  the  remainder  by  the 

ratio  less  1. 

NOTE. — Let  every  descending  series  be  inverted,  and  the  first  term  called  the 
last;  then  the  ratio  will  be  greater  than  a  unit.  If  the  series  be  infinite,  the  least 
term  is  a  cipher. 

EXAMPLES    FOR   PRACTICE. 

1.  The  extremes  are  3  and  384,  and  the  ratio  is  2;  what  is  the 
sum  of  the  series  ?  Ans.  765. 

2.  If  the  extremes  are  5  and  1080,  and  the  ratio  is  6,  what  is 
the  sum  of  the  series  ? 

3.  If  the  first  term  is  44,  the  last  term  ^|^,  and  the  ratio  £, 
what  is  the  sum  of  the  series  ?  Ans.   7^^. 

4.  What  is  the  sum  of  the  infinite  series,  8,  4,  2,  17  £,  i,  etc.? 

PROBLEM    V. 

686.  Given,  the  first  term,  the  ratio,  and  the  num- 
ber of  terms,  to  find  the  sum  of  the  series. 

If,  for  example,  the  first  term  be  4,  the  ratio  3,  and  the  number  of 
terms  6,  then  by  Problem  I,  we  have 

4  x  3*  ==  the  last  term. 


406  SERIES. 

Whence  by  Problem  IV,  we  nave 

4  X  36  —  4       (36  —  1)  X  4 

-  =.- v  — '• =  1456,  the  sum  of  the  series, 

3—1  3  —  1 

Hence  the  following 

RULE.  Raise  the  ratio  to  a  power  indicated  by  the  number  of 
terms,  and  subtract  1  from  the  result ;  then  multiply  this  remainder 
by  the  first  term^  and  divide  the  product  by  th?  ratio  less  1. 

EXAMPLES    FOR   PRACTICE. 

1.  The  first  term  is  7,  the  ratio  3,  and  the  number  of  terms  4 ; 
what  is  the  sum  of  the  series  ?  Arts.  280. 

2.  The  first  term  is  375,  the  ratio  J,  and  the  number  of  terms 
4 ;  what  is  the  sum  of  the  series  ? 

3.  The  first  term  is  175,  the  ratio  1.06,  and  the  number  of  terms 
5;  what  is  the  sum  of  the  series?  Ans.  986.49-f-- 

PROBLEM   VI. 

687.  Given,  the  extremes  and  the  sum  of  the  series, 
to  find  the  ratio. 

If  we  take  the  geometrical  progression,  2,  6,  18,  54,  162,  in  which 
the  ratio  is  3,  and  remove  the  first  term  and  the  last  term,  succes- 
sively, and  then  compare  the  results,  we  have 

6  -f  18  4-  54  4-  162  =  sum  of  the  series  minus  the  first  term. 

2  4-    6  ~|-  18  4-    54  =  sum  of  the  series  minus  the  last  term. 
Now,  since  every  term  in  the  first  line  is  3  times  the  corresponding 
term  in  the  second  line,  the  sum  of  the  terms  in  the  first  line  must 
be  3  times  the  sum  of  the  terms  in  the  second  line.     Hence  the 

RULE.  Divide  the  sum  of  the  series  minus  the  first  term,  by  the 
sum  of  the  series  minus  the  last  term. 

EXAMPLES    FOR    PRACTICE. 

1.  The  extremes  are  2  and  686,  and  the  sum  of  the  series  is 
800 ;  what  is  the  ratio  ?  Ans.  1. 


GEOMETRICAL  PROGRESSION.  407 

2.  The  extremes  are  J  and  64,  and   the  sum  of  the  series  is 
1-7|;  what  is  the  ratio? 

3.  If  the  sum  of  an  infinite  series  be  4£,  and  the  greater  ex- 
treme 3,  what  is  the  ratio  ?  Ans.  £. 

O88*  Every  other  problem  in  Geometrical  Progression,  that 
admits  of  an  arithmetical  solution,  may  be  solved  either  by  re- 
versing or  combining  some  of  the  problems  already  given. 

COMPOUND    INTEREST    BY    GEOMETRICAL   PROGRESSION. 

689.  We  have  seen  (55O)  that  if  any  sum  at  compound  in- 
terest be  multiplied  by  the  amount  of  $1  for  the  given  interval, 
the  product  will  be  the  amount  of  the  given  sum  or  principal  at 
the  end  of  the  first  interval  j  and  that  this  amount  constitutes  a 
new  principal  for  the  second  interval,  and  so  on  for  a  third,  fourth, 
or  any  other  interval.  Hence, 

A  question  in  compound  interest  constitutes  a  geometrical  pro- 
gression, whose  first  term  is  the  principal;  the  common  multiplier 
or  ratio  is  one  plus  the  rate  per  cent,  for  one  interval ;  the  number 
of  terms  is  equal  to  the  number  of  intervals  -|-1 ;  and  the  last 
term  is  the  amount  of  the  given  principal  for  the  given  time.  All 
the  usual  cases  of  compound  interest  and  discount  computed  at 
compound  interest,  can  therefore  be  solved  by  the  rules  for  geo- 
metrical progression.  For  example, 

Find  the  amount  of  $250  for  4  years,  at  6  %  compound 
interest. 

OPERATION. 

$250  x  1.06*  =  $250  x  1.262477  —  $316.21925. 

ANALYSIS.  Here  we  have  $250  the  first  term,  1.06  the  ratio,  and 
5  the  number  of  terms,  to  find  the  last  term.  Then  by  682  we  find 
the  last  term,  which  is  the  amount  required. 

EXAMPLES    FOR   PRACTICE. 

1.  What  is  the  amount  of  $350  in  4  years,  at  6  %  per  annum 
compound  interest  ?  Ans.  $441.86. 

2.  Of  what  principal  is  $150  the  compound  interest  for  2  years, 
at  7  %  ? 


408  SERIES. 

3.  What  sum  at  6  %  compound  interest,  will  amount  to  $1000 
in  3  years  ?  Ans.  $839.62. 

4.  In  how  many  years  will  $10  amount  to  $53.24,  at  10  %  com- 
pound interest?  Ans.  3  years. 

5.  At  what  rate  per  cent .  compound  interest  will  any  sum  double 
itself  in  8  years?  Ans.  9.05  -f  %. 

6.  What  is  the  present  worth  of  $322.51,  at  5  %  compound 
interest,  due  24  years  hence  ?  Ans.  $100. 


ANNUITIES. 

690.  An  Annuity  is  literally  a  sum  of  money  which  is  pay- 
able annually.  The  term  is,  however,  applied  to  a  sum  which  is 
payable  at  any  equal  intervals,  as  monthly,  quarterly,  semi-annu- 
ally,  etc. 

NOTE. — The  term,  interval,  will  be  used  to  denote  the  time  between  payments. 
Annuities  are  of  three  kinds :    Certain,  Contingent,  and  Per- 
petual. 

691*  A  Certain  Annuity  is  one  whose  period  of  continu- 
-  ance  is  definite  or  fixed. 

692.  A  Contingent  Annuity  is  one  whose  time  of  commence- 
ment, or  ending,  or  both,  is  uncertain ;  and  hence  the  period  of 
its  continuance  is  uncertain. 

693.  A  Perpetual  Annuity  or  Perpetuity  is  one  which  con- 
tinues forever. 

694:.  Each  of  these  kinds  is  subject,  in  reference  to  its  com- 
mencement, to  the  three  following  conditions : 

1st.  It  may  be  deferred,  i.  e.,  it  is  not  to  be  entered  upon  until 
after  a  certain  period  of  time. 

2d.  It  may  be  reversionary,  i.  e.,  it  is  not  to  be  entered  upon 
until  after  the  death  of  a  certain  person,  or  the  occurrence  of  some 
certain  event. 

3d.  It  may  be  in  possessionj  i.  e.,  it  is  to  be  entered  upon  at 
tmce. 


ANNUITIES.  409 

695.  An  Annuity  in  Arrears  or  Forborne  is  one  on  which 
the  payments  were  not  made  when  due.  Interest  is  to  be  reck- 
oned on  each  payment  of  an  annuity  in  arrears,  from  its  maturity, 
the  same  as  on  any  other  debt. 

ANNUITIES  AT  SIMPLE  INTEREST. 
GOG.   In  reference  to  an  annuity  at  simple  interest,  we  observe : 

I.  The  first  payment  becomes  due  at  the  end  of  the  first  inter- 
val, and  hence  will  bear  interest  until  the  annuity  is  settled. 

II.  The  second  payment  becomes  due  at  the  end  of  the  second 
interval,  and  hence  will  bear  interest  for  one  interval  less  than  the 
first  payment. 

III.  The  third  payment  will  bear  interest  for  one  interval  less 
than  the  second ;  and  so  on  to  any  number  of  terms.     Hence, 

IV.  All  the  payments  being  settled  at  one  time,  eacl*  will  be 
less  than  the  preceding,  by  the  interest  on  the  annuity  for  one 
interval.     Therefore,  they  will  constitute  a  descending  arithmetical 
progression,  whose  first  term  is  the  annuity  plus  its  interest  for  as 
many  intervals  less  one  as  intervene  between  the  commencement 
and  settlement  of  the  annuity;  the  common  difference  is  the  in- 
terest on  the  annuity  for  one  interval ;  the  number  of  terms  is  the 
number  of  intervals  between  the  commencement  and  settlement 
of  the  annuity;  and  the  last  term  is  the  annuity  itself. 

6H7*  The  rules  in  Arithmetical  Progression  will  solve  all 
problems  in  annuities  at  simple  interest. 

EXAMPLES    FOR   PRACTICE. 

1.  A  man  works  for  a  farmer  one  year  and  six  months,  at  $20 
per  month,  payable  monthly;  and  these  wages  remain  unpaid 
until  the  expiration  of  the  whole  term  of  service.  How  much  is 
due  to  the  workman,  allowing  simple  interest  at  6  per  cent,  per 
annum  ? 

OPERATION.  ANALYSIS.    Here  the 

$20  +  $.10  X  17  =  $21.70,  first  term.     last    month'8     wages, 

$20  -f  $21.70  $20>  is  the  last  term; 

- 2~      -  X  18  =  375.30,  sum.        the  number  of  months, 

18,  is  the   number  of 


410  SERIES. 

terms  ;  and  the  interest  on  1  month's  wages,  $.10,  is  the  common  dif- 
ference ;  and  since  the  first  month's  wages  has  been  on  interest  17 
months,  the  progression  is  a  descending  series.  Then,  by  677  we  find 
the  first  term,  which  is  the  amount  of  the  first  month's  wages  for  17 
months  ;  and  by  680  we  find  the  sum  of  the  series,  which  is  the  sum 
of  all  the  wages  and  interest. 

2.  A  father  deposits  annually  for  the  benefit  of  his  son,  com- 
mencing with  his  tenth  birthday,  such  a  sum  that  on  his  21st 
birthday  the  first  deposit  at  simple  interest  amounts  to  $210,  and 
the  sum  due  his  son  to  $1860.  How  much  is  the  deposit,  and  at 
what  rate  per  cent,  is  it  deposited  ? 

OPERATION.  ANALYSIS.  Here  the 

$1860  X  2—  $210  X  12  $210,  the  amount  of 

12  W'  deP°slt-     -    the   first  deposit,   is 


21Q  _  100  *^e  nrst   term  ;    12, 

—  jy  --  =  10  %,  fate-  the  number  of  depo- 

sits, is  the  number  of 

terms  ;  and  $1860,  the  amount  of  all  the  deposits  and  interests,  is  the 
sum  of  the  series.  By  680  we  find  the  last  term  to  be  $100,  which 
is  the  annual  deposit  ;  and  by  678  we  find  the  common  difference  to 
be  $10,  which  is  the  annual  rate  %  . 

3.  What  is  the  amount  of  an  annuity  of  $150  for  5£  years,  pay- 
able quarterly,  at  1J  per  cent,  per  quarter?         Ans.  $3819.75. 

4.  In  what   time  will  an   annual  pension  of  $500  amount  to 
$3450,  at  6  per  cent,  simple  interest  ?  Ans.  6  years. 

5.  Find  the  rate  per  cent  at  which  an  annuity  of  $6000  will 
amount  to  $59760  in  8  years,  at  simple  interest. 

Ans.  7  per  cent. 

ANNUITIES  AT  COMPOUND  INTEREST. 

OO8*  Ari  Annuity  at  compound  interest  constitutes  a  geomek 
rical  progression  whose  first  term  is  the  annuity  itself;  the  common 
multiplier  is  one  plus  the  rate  per  cent,  for  one  interval  expressed 
decimally;  the  number  of  terms  is  the  number  of  intervals  for  which 
the  annuity  is  taken  j  and  the  last  term  is  the  first  term  multiplied 
by  one  plus  the  rate  per  cent,  for  one  interval  raised  to  a  power 
one  less  than  the  number  of  terms. 


PROMISCUOUS  EXAMPLES.  4U 

699.  The  Present  Value  of  an  Annuity  is  such  a  sum  as 
would  produce,  at  compound  interest,  at  a  given  rate,  the  same 
amount  as  the  sum  of  all  the  payments  of  the  annuity  at  com- 
pound interest.  Hence,  to  find  the  present  value;  —  First  find  the 
amount  of  the  annuity  at  the  given  rate  and  for  the  given  time  ly 
68G5  then  find  the  present  value  of  this  amount  by 
taking  out  the  amount  of  $1,  or  divisor,  from 


NOTES.  —  1.  The  present  value  of  a  reversionary  annuity  is  that  principal  which 
will  nmount,  at  the  time  the  reversion  expires,  to  what  will  then  be  the  present 
vnlue  of  the  annuity. 

2.  The  present  value  of  a  perpetuity  is  a  sum  whose  interest  equals  the  an- 
nuity. 

TOO,  Questions  in  Annuities  at  compound  interest  can  be 
solved  by  the  rules  of  Geometrical  Progression. 


PROMISCUOUS    EXAMPLES    IN    SERIES. 

1.   Allowing  6  per  cent,  compound  interest  on  an  annuity  of 
$200  which  is  in  arrears  20  years,  what  is  its  present  amount  ? 

Ans.  $7357.11. 

'2.  Find  the  annuity  whose  amount  for  25  years  is  $16459.35, 
allowing  compound  interest  at  6  per  cent.  Ans.   $300. 

3.  What  is  the  present  worth  of  an  annuity  of  $500  for  7  years, 
at  6  per  cent,  compound  interest?  Ans.  $2791.18. 

4.  What  is  the  present  value  of  a  reversionary  lease  of  $100, 
commencing  14  years  hence,  and  to  continue  20  years,  compound 
interest  at  5  per  cent.?  Ans.  $629.426. 

5.  Find  the  sum  of  21  terms  of  the  series,  5,  4|,  4£,  etc. 

6.  A  man  traveled  13  days;  his  last  day's  journey  was  80  miles, 
and  each  day  he  traveled  5  miles  more  than  on  the  preceding  day. 
How  far  did  he  travel,  and  what  was  his  first  day's  journey? 

Ans.   He  traveled  650  miles. 

7.  Find  the  12th  term  of  the  series,  30,  15,  7£,  etc. 

Ans.  Tif?. 

8.  The  first  term  of  a  geometrical  progression  is  2,  the  last 
512,  and  common  multiplier  4 ;  find  the  sum  of  the  series. 

Ans.  682. 


412  SERIES. 

9.  The  distance  between  two  places  is  360  miles.    In  how  many 
days  can  it  be  traveled,  by  a  man  who  travels  the  first  day  27 
miles,  and  the  last  day  45,  each  day's  journey  being  greater  than 
the  preceding  by  the. same  number  of  miles?  Ans.   10. 

10.  The  first  term  of  a  geometrical  progression  is  1,  the  last 
term  15625,  and  the  number  of  terms  7j  find  the  common  ratio. 

Ans.  5. 

11.  An  annual  pension  of  $500  is  in  arrears  10  years.     What 
is  the  amount  now  due,  allowing  6  per  cent,  compound  interest  ? 

Ans.  $6590.40. 

12.  Find  the  first  and  last  terms  of  an  arithmetical  progression 
whose  sum  is  408,  common  difference  6,  and  number  of  terms  8. 

Ans.  First  term,  30 ;  last  term,  72. 

13.  A  farmer  pays  $1196,  in  13  quarterly  payments,  in  such  a 
way  that  each  payment  is   greater  than  the  preceding  by  $12. 
What  are  his  first  and  last  payments  ?         Ans.  $20,  and  $164. 

14.  A  man  wishes  to  discharge  a  debt  in  yearly  payments,  mak- 
ing the  first  payment  $2,  the  last  $512,  and  each   payment  four 
times  the  preceding  payment.     How  long  will  it  take  him  to  dis* 
charge  the  debt,  and  what  is  the  amount  of  his  indebtedness  ? 

15.  A  man  dying,  left  5  sons,  to  whom  he  gave  his  property  as 
follows :  to  the  youngest  he  gave  $4800,  and  to  each  of  the  otheri 
1 J  times  the  next  younger  son's  share.     What  was  the  eldest  son's 
fortune,  and  what  the  amount  of  property  left  ? 

Ans.  Eldest  son's  share,  $24300 ;  property,  $63300. 

16.  Find  the  annuity  whose  amount  for  5  years,  at  6  per  cent, 
compound  interest,  is  $2818.546.  Am.  $500. 

17.  A  merchant  pays  a  debt  in  yearly  payments  in  such  a  way 
that  each  payment  is  3  times  the  preceding ;  his  first  payment  is 
$10,  and  his  last  $7290.     What  is  the  amount  of  the  debt,  and  in 
how  many  payments  is  it  discharged  ? 

Ans.  Debt,  $10930;  7  payments. 

18.  A  man  traveling  along  a  road,  stopped  at  a  number  of 
stations,  but  at  each  station  he  found  it  necessary,  before  proceed- 
ing to  the  next,  to  return  to  the  place  from   which  he  first  started  j 


PROMISCUOUS  EXAMPLES.  413 

the  distance  from  the  starting  place  to  the  first  station  was  D  miles, 
and  to  the  last  25  mile's;  he  traveled  in  all  180  miles.  How 
many  stations  were  there  on  the  road,  and  what  was  the  distance 
from  station  to  station  ?  Ans.  6  stations ;  4  miles  apart. 

19.  An  annuity  of  §200  for  12  years  is  in  reversion  6  years 
What  is  its  present  worth,  compound  interest  at  6  %  ? 

Ans.  $1182.05  +  . 

20.  A  man  pays  $6  yearly  for  tobacco,  from  the  age  of  16  until 
he  is  60,  when  he  dies,  leaving  to  his  heirs  $500.     What  might 
he  have  left  them,  if  he  had  dispensed  with  this  useless  habit  and 
loaned   the   money  at  the  end  of  each  year  at  6   %   compound 
interest?  Ans.  $1698.548+. 

21.  What  is  the  present  worth  ot  a  reversionary  perpetuity  of 
$100,  commencing  30  years  hejipe,  allowing  5  per  cent,  compound 
interest?  An?.  $462.75-}-. 

22.  Two  boys,  each  12  years  old,  have  certain  sums  of  money 
left  to  them ;  the  sum  left  to  one  is  put  out  at  V  %  simple  inte- 
rest, and  the  sum  left  the  other  at  6  %  compouiiC  interest,  paya- 
ble semi-annually,  and  the  amount  cf  each,  boy's  money  will  be 
$2000  when  he  is  21  years  old.     What  is  the  sum  left  to  each 
boy? 

23.  A  merchant  purchased  8  pieces  of  cloth,  for  which  he  paid 
$136 ;  the  difference  in  the  length  of  any  two  pieces  was  2  yds.^ 
and  the  difference  in  the  price  $4.     He  paid  $31  for  the  longest 
piece,  and  $1  a  yard  for  the  shortest.     Find  the  whole  number  of 
yards,  and  the  price  per  yard  of  each  piece. 

24.  A  farmer  has  600  bushels  of  different  kinds  of  grain,  mixed 
in  such  a  way  that  the  number  of  bushels  of  the  several  kinds  con- 
stitute a  geometrical  progression,  whose  common  multiplier  is  2 ; 
the  greatest  number  of  bushels  of  one  kind  is  320.     Find  the 
number  of  kinds  of  grain  in  the  mixture,  and  the  number  of 
bushels  of  each  kind.  Ans.  4  kinds. 


MISCELLANEOUS    EXAMPLES. 


MISCELLANEOUS  EXAMPLES. 

1.  How  many  thousand  shingles  will  cover  both  sides  of  a 
roof  36  ft.  long,  and  whose  rafters  are  18  ft.  in  length  ? 

2.  From  f  of  4  of  £  of  70  miles,  subtract  .73  of  1  mi.  3  fur. 
•     3.  What  number  is  that  from  which  if  7£  be  subtracted,  f  of  the 
remainder  is  91£  ?  Ans.  144|. 

4.  What  part  of  4  is  £  of  6?  Ans.  f. 

5  It  is  required  to  mix  together  brandy  at  $.80  a  gallon,  wine  at 
$.70,  cider  at  $.10,  and  water,  in  such  proportions  that  the  mixture 
may  be  worth  $.50  a  gallon ;  what  quantity  of  each  must  be  used  ? 

Ans.  3  gal  of  water,  2  of  cider,  4  of  wine,  and  5  of  brandy. 

6.  What  number  increased  by  J,  J,  and  J  of  itself  equals  125  ? 

7.  What  is  the  hour,  when  the  time  past  noon  is  equal  to  f  of  the 
time  to  midnight  ?  Ans.  4  h.  48  min.  p.  M. 

8.  A  grocer  mixed  12  cwt.  of  sugar  @  $10,  with  3  cwt.  @  $8f,  and 
8  cwt.  @  $7£ ;  how  much  was  1  cwt.  of  the  mixture  worth  ? 

9.  If  $240  gain  $5.84  in  4  mo.  26  da.,  what  is  the  rate  <f0  ?   Ans.  6. 

10.  If  24  men,  in  189  da.,  working  10  h.  a  day,  dig  a  trench  33|  yd. 
long,  2|  yd.  deep,  and  5]  yd.  wide;  how  many  hoursji  day  must  217 
men  work,  to  dig  a  trench  23 }  yd.  long,  2£  yd.  deep,  and  3|  yd.  wide, 
in  5J  days?  Ans.  16  h. 

11.  What  is  the  difference  between  the  interest  and  the  discount  of 
$450  at  5  per  cent.,  for  6  yr.  10  mo.  ? 

12.  A  younger  brother  received  $6300,  which  Avas  £  as  much  as  his 
elder  brother  received;  how  much  did  both  receive? 

13.  Keduce  .7,  .88,  .727,  .91325  to  their  equivalent  common  fractions. 

14.  A  person  by  selling  a  lot  of  goods  for  $438,  loses  10  % ;  how 
much  should  the  goods  have  been  sold  for,  to  gain  12£  ft  ? 

15.  For  what  sum  must  a  note  be  drawn  at  4  mo.,  that  the  proceeds 
of  it,  when  discounted  at  bank  at  7  per  cent.,  shall  be  $875.50? 

16.  Three  persons  engaged  in  trade  with  a  joint  capital  of  $2128; 
A's  capital  was  in  trade  5  mo.,  B's  8  mo.,  and  C's  12  mo.;  A's  share 
of  the  gain  was  $228,  B's  $266.40,  and  C's  $330.    What  was  the  capital 
of  each?  Ans.  A's,  $912;  B's,  $666;  C's,  $555. 

17  Henry  Truman  purchased  corn  of  John  Bates,  on  2  months' 
credit,  as  follows:  Aug.  27,  300  bu.  @  $.35;  Aug.  31, 150  bu.  @  $.40; 
Sept.  7,  500  bu.  @  $.38;  Sept.  12,  200  bu.  @  $.42;  Sept.  25,  250  bu. 
@  $.40.  When  was  the  a|c  due  per  average  ? 

18.  A  B  and  C  can  do  a  job  of  work  in  12  da.,  C  can  do  it  in  24  da., 
and  A  in  34  da. ;  in  what  time  can  B  do  it  alone?        Ans.  812  da. 

19.  If  a  man  travel  7  mi.  the  first  day,  and  51  mi.  the  last,  increas- 
ing his  iourney  4  mi.  each  day,  how  many  days  will  he  travel,  and 
how  far  ?  Ans.  12  da.,  and  348  mi. 


MISCELLANEOUS   EXAMPLES.  415 

20.  What  is  the  difference  between  the  true  and  bank  discount  of 
$2500,  payable  in  90  days  at  7  per  cent.?  Ans,  $2.21. 

21.  Which  is  the  more  advantageous,  to  buy  flour  at  $5  a  bbl.  on  6 
mo.,  or  $4.87 £  cash,  money  being  worth  7  <fi>  ?      Ans.  At  $5  on  6  mo. 

22.  Sold  J  of  a  lot  of  lumber  for  what  f  of  it  cost ;  what  fo  was 
gained  on  the  part  sold?  Ans.  25  ^,. 

23.  If  $500  gain  $50  in  1  yr.,  in  what  time  will  $960  gain  $60? 

24.  Received  an  invoice  of  crockery,   12  per  cent,  of  which  was 
broken  ;  at  what  per  cent,  above  cost  must  the  remainder  be  sold,  to 
clear  25  per  cent,  on  the  invoice?  Ans.  ^2^. 

25.  The  sum  of  two  numbers  is  365,  and  their  difference  is  .0675 ; 
what  are  the  numbers  ? 

26.  If  the  interest  of  $445.62*  be  $128.99  for  7  yr.,  what  will  be  the 
interest  of  $650  for  3  yr.  10  mo.  15  da.  ? 

27.  Received  from  Savannah  150  bales  of  cotton,  each  weighing 
540  lb.,  and  invoiced  at  7d.  a  pound  Georgia  currency.     Sold  it  at  an 
advance  of  26  ^,,  commission  lj  ^,  and  remitted  the  proceeds  by 
draft.    What  was  the  face  of  the  draft,  exchange  being  £  c/c  discount? 

Ans.  $12629.28+. 

28.  A  man  in  Chicago  has  5000  francs  due  him  on  account  in  Paris, 
lie  can  draw  on  Paris  for  this  amount,  and  negotiate  the  bill  at  19  f 
cents  per  franc ;  or  he  can  advise  his  correspondent  in  Paris  to  remit 
a  draft  on  the  United  States,  purchased  with  the  sum  due  him,  ex- 
change on  U.  S.  being  at  the  rate  of  5  francs  20  centimes  per  $1. 
What  sum  will  the  man  receive  by  each  method? 

Ans.  By  draft  on  Paris,  $970 ;  by  remittance  from  Paris,  $961.53. 

29.  What  sum  must  be  invested  in  stocks  bearing  6^  </c,  at  105$, to 
produce  an  income  of  $1000?  Ans.  $16153.84. 

30.  A  person  exchanges  250  shares  of  6  per  cent,  stock,  at  70  <&, 
for  stock  bearing  8  per  cent.,  at  120  <J0 ;  what  is  the  difference  in  his 
income?  Ans.  $333.33J. 

31.  If  f  of  A's  money  equals  f  of  B's,  and  f  of  B's  equals  j  of  C's, 
and  the  interest  of  all  their  money  for  4  yr.  8  mo.  at  6  ft  is  $15190, 
how  much  money  has  each  ? 

Ans.  A  has  $18859.44+ ;  B,  $16763.95+;  C,  $18626.61. 

32.  A  boy  14  years  old  is  left  an  annuity  of  $250,  which  is  de- 
posited in  a  savings  bank  at  6  <&,  interest  payable  semi-annually ; 
now  much  will  he  be  worth  when  of  age?  Ans.  $2104.227. 

33.  If  a  boy  buys  peaches  at  the  rate  of  5  for  2  cents,  and  sells 
them  at  the  rate  of  4  for  3  cents,  how  many  must  he  buy  and  sell  to 
make  a  profit  of  $4.20  ? 

34.  \Yhat  fo  in  advance  of  the  cost  must  a  merchant  mark  his 
goods,  so  that,  after  allowing  5  (f0  of  his  sales  for  bad  debts,  an  ave- 
rage credit  of  6  months,  and  7  <fo  of  the  cost  of  the  goods  for  his  ex- 
penses, he  may  make  a  clear  gain  of  12£  ft  on  the  first  cost  of  the 
goods,  money  being  worth  6  <fc  ?  Ans.  29.56 -f  <fc. 


416  MISCELLANEOUS  EXAMPLES. 

35.  Four  men  contracted  to  do  a  certain  job  of  work  for  $8600;  the 
first  employed  28  laborers  20  da.,  10  h.  a  day;  the  second,  25  laborers 
15  da.,  12  h.  a  day ;  the  third,  18  laborers  25  da.,  11  h.  a  day ;   and 
the  fourth,  15  laborers  24  da.,  8  h.  a  day.     How  much  should  each 
contractor  receive  ? 

Ans.  1st,  $2686;  2d,  $2158.39;  3d,  $2374.24;  4th,  $1381.37. 

36.  If  I  exchange  75  railroad  bonds  of  $500  each,  at  36  %  below 
par,  for  bank  sjjock  at  5  %  premium,  how  many  shares  of  $100  each 
will  I  receive?  Ans.  228^. 

37.  A  trader  has  bought  merchandise  as  follows  :  July  3,  $35.26  ; 
'July  4,  $48.65,  on  30  da.;  Aug.  17,  $6.48;  Sept.  12,  $50.     What  is 
due  on  the  account  Oct.  12,  interest  at  9  %  ?  Ans.  142.60. 

38.  A  farmer  sold  34  bu.  of  corn,  and  56  bu.  of  barley  for  $63.10, 
receiving  35  cents  a  bushel  more  for  the  barley  than  ibr  the  corn ; 
what  was  the  price  of  each  per  bushel? 

39.  A  speculator  purchased  a  quantity  of  flour,  Sept.  1 ;  Oct.  1  its 
value  had  increased  25  %  ;  Nov.  1  its  value  was  30  %  more  than  Oct. 
1;  Dec.  1  he  sold  it  for  15  %  less  than  its  value  Nov.  1,  receiving  in 
payment  a  6  months'  note,  which  he  got  discounted  at  a  bank,  at  7 
Jo,  receiving  $12950  on  it.     How  much  was  his  profit  on  the  flour? 

Ans.  $3228.51. 

40.  A  flour  merchant  bought   120  bbl.  of  flour  for  $660,  paying 
$5.75  for  first  quality  and  $5  for  second  quality ;  how  many  barrels 
were  first  quality  ?  Ans.  80. 

41.  Two  mechanics  work  together ;  for  15  days'  work  of  the  first 
and  8  days'  work  of  the  second  they  receive  $61,  and  for  6  days' 
work  of   the  first  and  10  days'  work    of  the   second    they  receive 
$38 ;  how  much  does  each  man  earn  ?  Ans.  1st,  $63  ;  2d,  $36. 

42.  The  duty,  at  15  %,  on  Rio  coffee,  in  bags  weighing  180  Ibs. 
gross,  and  invoiced  at  $.12J  per  pound,  was  $961.87^,  tare  having  been 
allowed  at  5  %  ;  how  many  bags  were  imported  ?  Ans.  300. 

43.  A  dairyman  took  some  butter  to  market,  for  which  he  received 
$49,  receiving  as  many  cents  a  pound  as  there  were  pounds ;  how 
many  pounds  were  there  ?  Ans.  70  Ib. 

44.  A  mechanic  received  $2  a  day  for  his  labor,  and  paid  $4  a  week 
for  his  board  ;  at  the  expiration  of  10  weeks  he  had  saved  $72 ;  how 
many  days  did  he  work,  and  how  many  was  he  idle  ? 

45.  To  what  would  $250,  deposited  in  a  savings  bank,  amount  in 
10  yr.,  interest  being  allowed  semi-annually  at  6  %  per  annum  ? 

46.  How  much  water  is  there  in  a  mixture  of  100  gal.  of  wine  anJ. 
wa,ter,  worth  $1  per  gal.,  if  100  gal.  of  the  wine  cost  $120  ? 

47.  If  a  pipe  3  in.  in  diameter  will  discharge  a  certain  quantity  of 
water  in  2  h.,  in  what  time  will  3  two-inch  pipes  discharge  3  times 
the  quantity?  Ans.  4  h.  30  min. 

48.  Wm,  Jones  &  Co.   become  insolvent  and  owe  $8100.     Their 
assets  amount  to  $4981.50.     What  per  cent,  of  their  indebtedness  can 


MISCELLANEOUS  EXAMPLES.  417 

they  pay,  allowing  the  assignees  2£  % '  on  the  amount  distributed 
for  their  services?  Ans.  GO  per  cent. 

49.  Shipped  a  car  load  of  fat  cattle  to  Boston,  and  offered  them  for 
sale  at  25  per  cent,  advance  on  the  cost :  but  the  market  being  dull  I 
sold  for  14  per  cent,  less  than  my  asking  price,  and  gained  thereby 
$170.     How  much  did  the  cattle  cost ;  for  how  much  did  they  sell ; 
and  what  was  my  asking  price? 

Ans.  Cost  $2266.66| ;  sold  for  $2436.66f ;  asking  price,  $2833.33 £. 

50.  What  must  be  the  dimensions  of  a  cubical  cistern  to  hold  2000 
gallons  ? 

51.  A  man  died  leaving  $5000  to  be  divided  between  his  three  sons, 
aged  13,  15,  and  16  yr.  6  mo.,  respectively,  in  such  a  proportion  that 
the  share  of  each  being  put  at  simple  interest  at  6  % ,  should  amount 
to  the  same  sum  when  they  should  arrive  at  the  age  of  21.     How 
much  was  each  one's  share  ? 

Ans.  Youngest,  $1536.76+  ;  second,  $1672.364- ;  oldest,  $1790.88  +  . 

52.  A  vessel  having  sailed  due  south  and  due  east  on  alternate  days, 
was  found,  after  a  certain  time,  to  be  118.794  miles  south-east  of  the 
place  of  starting  ;  what  distance  had  she  sailed  ?     Ans.  168  miles. 

53.  Imported  4  pipes  of  Madeira  wine,  at  §2.15  a  gallon,  and  paid 
$57.60  freight,  and  a  duty  of  24  per  cent.    I  sold  the  whole  for  $1980 ; 
what  was  my  gain  °fo  ? 

54.  If  34J  bu.  of  corn  are  equal  in  value  to  17  bu.  wheat,  9  bu.  of 
wheat  to  59£  bu.  of  oats,  and  6  bu.  of  oats  to  42  Ib.  of  flour,  how  many 
bushels  of  corn  will  purchase  5  bbl.  of  flour  ?  Ans.  42f£f. 

55.  If  stock  bought  at  8  %  discount  will  pay  7  %  on  the  invest- 
ment, at  what  rate  should  it  be  bought  to  pay  10  %  ? 

56.  A  merchant  in  New  York  gave  $2000  for  a  bill  of  exchange  of 
£400  to  remit  to  Liverpool ;  what  was  the  rate  in  favor  of  England  ? 

57.  A,  B,  and  C  start  from  the  same  point,  to  travel  around  a  lake 
84  miles  in  circumference.     A  travels  7  miles,  and  B  21  miles  a  day 
in  the  same  direction,  and  C  14  miles  in  an  opposite  direction.     In 
how  many  days  will  they  all  meet  ?  Ans.  12. 

58.  The  exact  solar  year  is  greater  than  365  days  by  {\$$&  of  a 
day  ;  find  approximately  how  often  leap  year  should  come,  or  one  day 
be  added  to  the  common  year,  in  order  to  keep  the  calendar  right  ? 

Ans.  Once  in  every  4  yr. ;  7  times  in  every  28  yr. ;  8  times  in  every 
33  yr. ;  31  times  in  every  128  yr. ;  or  163  times  in  every  673  yr. 

59.  A  gentleman  purchases  a  farm  for  $10000,  which  he  sells  after 
a  certain  number  of  years  for  $14071,  making  on  the  investment  5  °/o 
compound  interest.    He  now  invests  his  money  in  a  perpetuity,  which 
is  in  reversion  11  years  from  the  date  of  purchasing  the  farm.     A1-. 
lowing  6  %  compound  interest  for  the  use  of  money,  find  the  annuity 
and  the  length  of  time  he  owns  the  farm. 

Ans.  Annuity,  $1065.85 ;  owned  the  farm  7  yr. 


418  MISCELLANEOUS  EXAMPLES. 

60.  What  will  I  gain  %  by  purchasing  goods  on  6  mo.,  and  selling 
them  immediately  for  cash  at  cost,  money  being  worth  7  %  ? 

61.  What  sum  must  a  man  save  annually,  commencing  at  21  years 
of  age,  to  be  worth  $30000  when  he  is  50  years  old,  his  savings  being 
invested  at  5  %  compound  interest?  Ans.  $481.37. 

62.  Three  persons  are  to  share  $10000  in  the  ratio  of  3,  4,  and  5, 
but  the  first  dying  it  is  required  to  divide  the  whole  sum  equitably  be- 
tween the  other  two.     What  are  the  shares  of  the  other  two  ? 

Ans.  $4444*,  and  $5555  jf. 

63.  If  50  bbl.  of  flour  in  Chicago  are  worth  125  yd.  of  cloth  in  New 
York,  and  80  yd.  of  cloth  in  New  York  are  worth  6  bales  of  cotton  in 
Charleston,  and  13  bales  of  cotton  in  Charleston  are  worth  3£  hhd. 
of  sugar  in  New  Orleans,  how  many  hhd.  of  sugar  in  New  Orleans 
are  worth  1500  bbl.  of  flour  in  Chicago?  Ans.  75,^. 

64.  Seven  men  all  start  together  to  travel  the  same  way  round  an 
island  120  miles  in  circumference,  and  continue  to  travel  until  they 
all  come  together  again.     They  travel  5,  6£,  7£,  8£,  9£,  10£  and  11| 
miles  a  day  respectively.    In  how  many  days  will  they  all  be  together 
again?  Ans.  1440  ^da. 

65.  There  are  two  clocks  which  keep  perfect  time  when  their  pen- 
dulums beat  seconds.    The  first  loses  20  seconds  a  day,  and  the  second 
gains  15  seconds  a  day.     If  the  two  pendulums  beat  together  when 
both  dials  indicate  precisely  12  o'clock,  what  time  does  each  clock 
show  when  the  pendulums  next  beat  in  concert  ? 

Ans.  The  first  shows  41  min.  8  sec.  past  12 ;  and  the  second  41 
min.  9  sec.  past  12. 

66.  If  a  body  put  in  motion  move  ^  of  an  inch  the  first  second  of 
time,  1  in.  the  second  sec.,  3  in.  the  third,  and  so  continue  to  increase 
in  geometrical  ratio,  how  far  would  it  move  in  30  seconds  ? 

Ans.  541590730f !>fi  mi. 

67.  If  stock  bought  at  5  <fo  premium  will  pay  6  fo  on  the  invest- 
ment, what  fo  will  it  pay  if  bought  at  15  <fo  discount  ?    Ans.  7-^  %, 

68.  If  6  apples  and  7  peaches  cost  33  cts.,  and  10  apples  and  8 
peaches  cost  44  cts.,  what  is  the  price  of  one  of  each  ? 

Ans.  Apples,  2  cts. ;  peaches,  3  cts. 

69.  A  gentleman  in  dividing  his  estate  among  his  sons  gave  A  $9 
as  often  as  B  $5,  and  C  $3  as  often  as  B  $7.    C's  share  was  $3862.50; 
what  was  the  value  of  the  whole  estate?  Ans.  $29097.50. 

70.  A  farmer  sold  16  bu.  of  corn  and  20  bu.  of  rye  for  $30,  and  24 
bu.  of  corn  and  10  bu.  of  rye  for  $27.     How  much  per  bushel  did  he 
receive  for  each  ?  Ans.  Corn,  $.75  ;  rye,  $.90. 

71.  A  drover  sold  some  oxen  at  $28,  cows  at  $17,  and   sheep  at 
$7.50  per  head,  and  received  $749  for  the  lot.     There  were  twice  as 
many  cows  as  oxen,  and  three  times  as  many  sheep  as  cows.     How 
many  were  there  of  each  kind  ? 

72.  For  what  sum  must  a  vessel,  valued  at  $25000,  be  insured,  so 


MISCELLANEOUS  EXAMPLES.  419 

that  in  case  of  its  loss,  the  owners  may  recover  both  the  value  of  the 
vessel  and  the  premium  of  24  <fo  ? 

73  A  boy  hired  to  a  mechanic  for  20  weeks,  on  condition  that  he 
should  receive  $20  and  a  coat.     At  the  end  of  12  weeks  the  boy  quit 
work   when  it  was  found  that  he  was  entitled  to  $9  and  the  coat  ; 
what'  was  the  value  of  the  coat?  ATM.  $7.50. 

74  An  irregular  piece  of  land,  containing  540  A.  36  P.,  is  ex- 
changed for  a°square  piece   containing  the  same  area;  what  is  the 
length  of  one  of  its  sides  ?     If  divided  into  42  equal  squares,  what 
wilfbe  the  length  of  the  side  of  each? 

75  What  will  be  the  difference  in  the  expense  of  fencing  two  fields 
of  25  acres  each,  one  square,  and  the  other  in  the  form  of  a  rectangle, 
whose  length  is'twice  its  breadth,  the  fence  costing 


76.  At  what  time  between  5  and  6  o'clock  are  the  hour  and  minute 
hands  of  a  watch  exactly  together  ? 

77.  A  general,  forming  his  army  into  a  square,  had  284  men  re- 
maining ;  but  increasing  each  side  by  one  man,  he  wanted  25  men  to 
complete  the  square,     llow  many  men  had  he  ?  Ans.  24000. 

78.  Divide  $3648  among  3  persons,  so  that  the  share  of  the  first  to 
that  of  the  second  shall  be  as  7  to  9,  and  of  the  first  to  the  third  as  3 
to  4.  Ans.  $1008,  $1296,  $1344. 

79.  If  a  lot  of  land,  in  the  form  of  an  oblong  or  rectangle,  con- 
tains 6  A.  132  P.,  and  its  length  is  to  its  width  as  21  to  13,  what  are 
its  dimensions  ;  and  how  many  rods  of  fence  will  be  required  to  in- 
close it  ?  Ans,  to  last,  136  rd.  of  fence. 

80.  Five  persons  are  employed  to  build  a  house.     A,  B,  C,  and  D 
can  build  it  in  13  days  ;  A,  B,  C,  and  E  in  15  days  ;  A,  B,  D,  and  E 
in  12  days  ;  A,  C,  D,  and  E  in  19  days  ;  and  B,  C,  D,  and  E  in  14 
days.     In  how  many  days  can  all  together  build  it  ;  and  which  one 
could  do  the  work  alone  in  the  shortest  time  ? 

Ans.  HyViVf  da.  ;  B  in  shortest  time. 

81.  Divide  $500  among  3  persons,  in  such  a  manner  that  the  share 
of  the  second  may  be  £  greater  than  that  of  the  first,  and  the  share 
of  the  third  £  greater  than  that  of  the  second. 

Ans.  1st,  $105T5ff;  2d,  $157j^;  3d,  $236}f. 

82.  A  and  B  engage  in  trade  ;  A  puts  in  $5000,  and  at  the  end  offc 
4  mo.  takes  out  a  certain  sum.     B  puts  in  $2500,  and  at  the  end  of  5 
mo.  puts  in  $3000  more.     At  the  end  of  the  year  A's  gain  is  $1066§> 
and  B's  is  $1333^.     What  sum  did  A  take  out  at  the  end  of  4  mo.  ? 

Ans.  $2400. 

83.  What  sum  of  money,  with  its  semi-annual  dividends  of  5   fo 
invested  with  it,  will  amount  to  $12750  in  2  yr.  ?  Ans.  $10489.450-. 

84.  If  a  speculator  invests  $1500  in  flour,  and  pays  5  fo  for  freights, 
2  fo  for  commission,  and  the  flour  sells  at  20  fo  advance  on  cost  price, 
on  a  credit  of  90  days,  and  he  gets  this  paper  discounted  at  bank  at 
7  %  ,  and  repeats  the  operation  every  15  days,  investing  all  the  pro- 
ceeds each  time,  how  much  will  be  his  whole  gain  in  two  months  ? 


420  MISCELLANEOUS  EXAMPLES. 

85.  If  a  piece  of  silk  cost  $.80  per  yard,  at  what  price  shall  it  be 
marked,  that  the  merchant  may  sell  it  at  10  %  less  than  the  marked 
price,  and  still  make  20  %  proiit?  Ans.  $1.06f. 

86.  A  merchant  bought  20  pieces  of  cloth,  each  piece  containing 
25  yd.  at  $4f  per  yard  on  a  credit  of  9  mo. ;  he  sold  the  goods  at 
$4£  per  yard  on  a  credit  of  4  mo.     What  was  his  net  cash  gain, 
money  being  worth  6  ft  ?  Ans.  $173-85. 

87.  A  owes  B  $1200,  to  be  paid  in  equal  annual  payments  of  $200 
each ;  but  not  being  able  to  meet  these  payments  at  their  maturities, 
and  having  an  estate  10  years  in  reversion,  he  arranges  with  B  to 
wait  until  he  enters  upon  his  estate,  when  he  is  to  pay  B  the  whole 
amount,  with  8  %  compound  interest.     What  sum  will  B  then  re- 
ceive? Ans.  $1996.074+. 

88.  A  gentleman  who  was  entitled  to  a  perpetuity  of  $3000  a  year, 
provided  in  his  will  that,  after  his  decease,  his  oldest  son  should  receive 
it  for  10  yr.,  then  his  second  son  for  the  next  10  yr.,  and  a  literary 
institution  for  ever  afterward.     What  was  the  value  of  each  bequest 
at  the  time  of  his  decease,  allowing  compound  interest  at  6  fo  ? 

Ans.  To  oldest  son,  $22080.28 ;  to  second  son,  $12329.51 ;  to  insti- 
tution, $15590.23. 

89.  B  has  3  teams  engaged  in  transportation ;  his  horse  team  can 
perform  the  trip  in  5  days,  the  mule  team  in  7  days,  and  the  ox  team 
in  11  days.     Provided  they  start  together,  and  each  team  rests  a  day 
after  each  trip,  how  many  days  will  elapse  before  they  all  rest  the 
same  day?  Ans.  23  days. 

90.  A  man  bought  a  farm  for  $4500,  and  agreed  to  pay  principal 
and  interest  in  4  equal  annual  installments;  how  much  was  the  annual 
payment,  interest  being  6  fi  ?  Ans.  $1298.67  +  . 

91.  A  bought  a  piece  of  property  of  B,  and  gave  him  his  bond  for 
$6300,  dated  Jan.  1,  1860,  payable  in  6  equal  annual  instalments  of 
$1050,  the  first  to  be  paid  Jan.  1,  1861.     A  took  up  his  bond  Jan. 
1,  1864,  semi-annual  discount  at  the  rate  of  6  %  per  annum  on  the 
two  payments  which  fell  dne  alter  Jan.  1,  1864,  being  deducted ; 
what  sum  canceled  the  bond?  Ans.  $2972.54  +  . 

92.  A  gentleman  desires  to  set  out  a  rectangular  orchard  of  864  trees, 
go  plfrcad  that  the  number  of  rows  shall  be  to  the  number  of  trees  in  a 
STOW,  as  3  to  2.     If  the  trees  are  7  yards  apart,  how  much  ground  will 
the  orchard  occupy  ?  Ans.  39445  sq.  yd. 

93.  S.  C.  Wilder  bought  25  shares  of  bank  stock  at  an  advance  of 

6  %  on  the  par  value  of  $100.     From  the  time  of  purchase  until 
the  end  of  3  yr.  3  mo.  he  received  a  semi-annual  dividend  of  4  <fo, 
when  he  sold  the  stock  at  a  premium  of  11  </o.     Money  being  worth 

7  'fo  compound  interest,  how  much  did  he  gain?          Ans.  $137.31. 


DEFINITIONS. 


421 


Horizontal. 


MENSURATION. 

7  O2.  Mensuration  is  the  process  of  finding  the  number  of  units 
in  extension. 

LINES. 

703.  A  Straight  Line  is  a  line  that  does     

not  change  its  direction.     It  is  the  shortest 

distance  between  two  points. 

704.  A  Curved  Line  changes  its  direc- 
tion at  every  point. 

705.  Parallel  Lines  have  the  same  direc- 
tion ;  and  being  in  the  same  plane  and  equally 
distant  from  each  other,  they  can  never  meet. 

706.  A  Horizontal  Line  is  a  line  parallel 
to  the  horizon  or  water  level. 

707.  A  Perpendicular  Line  is  a  straight 
line  drawn  to  meet  another  straight  line,  so  as 
to  incline  no  more  to  the  one  side  than  to  the 
other. 

A  perpendicular  to  a  horizontal  Line  is  called  a 
vertical  line. 

708.  Oblique  Lines  approach  each  other, 
and  will  meet  if  sufficiently  extended. 

ANGLES. 

7®@.  An  Angle  is  the  opening  between 
two  lines  that  meet  each  other  in  a  common 
point,  called  the  vertex. 

Angles  are  measured  by  degrees  (3O1). 

710.  A  Right  Angle  is  an  angle  formed 
by  two1  lines  perpendicular  to  each  other. 

711.  An  Obtuse  Angle  is  greater  than  V 

a  right  angle.  \ 

712.  An  Acute  Angle  is  less  than  a  right  angle. 
All  angles  except  right  angles  are  called  oblique  angles. 


422 


MENSURATION. 


PLANE     FIGURES. 

713.  A  Plane  Figure  is  a  portion  of  a  plane  surface  bounded 
by  straight  or  curved  lines. 

:.  A  Polygon  is  a  plane  figure  bounded  by  straight  lines. 
.  The  Perimeter  of  a  polygon  is  the  sum  of  its  sides. 

716.  The  Area  of  a  plane  figure  is  the  surface  included  within 
the  lines  which  bound  it 


A  regular  polygon  has  all  its  sides  and  all  its  angles  equal. 
A  polygon  of  three  sides  is  called  a  trigon,  or  triangle  ;  of  four  sides, 
tetragon,  or  quadrilateral  ;  of  five  sides,  a  pentagon,  etc. 


Pentagon.       Hexagon.         Heptagon.        Octagon.          Nonagon.         Decagon. 


TRIANGLES. 

717.  A   Triangle   is  a  plane  figure  bounded  by  three  sides, 
and  having  three  angles. 

718.  A  Right- Angled  Triangle  is  a 

triangle  having  one  right  angle. 

719.  The   Hypothenuse    of    a  right- 
angled   triangle   is   the  side  opposite  the 

right  angle.  Base. 

720.  The  Base  of  a  triangle,  or  of  any  plane  figure,  is  the  side 
on  which  it  may  be  supposed  to  stand. 

731 .  The  Perpendicular  of  a  right-angled  triangle  is  ttoe  side 
which  forms  a  right  angle  with  the  base. 

7258.  The  Altitude  of  a  triangle  is  a  line  drawn  perpendicular 
to  the  base  from  the  angle  opposite. 

1.  The  dotted  vertical  lines  in  the  figures  represent  the  altitude. 

2.  Triangles  are  named  from  the  relation  both  of  their  sides  and  angle* 


TEIANGLES.  423 

733.  An  Equilateral  Triangle  has  its  three  sides  equal. 
724:.  An  Isosceles  Triangle  has  only  two  of  its  sides  equal. 
735.  A  Scalene  Triangle  has  all  of  its  sides  unequal. 

FIG.  1.  FIG.  2.  FIG.  3. 


Equilateral.  Isosceles.  Scalene. 


.  An  Equiangular  Triangle  has  three  equal  angles.  (Fig.  1.) 
727.  An  Acute-angled  Triangle  has  three  acute  angles.  (Fig.  2.) 

738.  An  Obtuse-angled  Triangle  has  one  obtuse  angle.  (Fig.  3.) 

PROBLEMS. 

739.  The  base  and  altitude  of  a  triangle  being  given  to 
find  its  area. 

1.  Find  the  area  of  a  triangle  whose  base  is  26  ft.  and  altitude 
14.5  ft. 

OPERATION.—  14.5  x  26-*-2=188isq.  ft.   Or,  26  x  -^-=188|  sq.  ft.,  area. 

S3 

2.  What  is  the  area  of  a  triangle  whose  altitude  is  1®  yd.  and 
base  40  ft.  ?  Ans.  600  sq.  ft. 

RULE  1.     Divide  the  product  of  the  base  and  altitude  by  2.     Or, 

2.  Multiply  the  base  by  one-half  the  altitude. 

Find  the  area  of  a  triangle 

3.  Whose  base  is  12  ft,  6  in.  and  altitude  6  ft,  9  in.   A.  42^8.^ 

4.  Whose  base  is  25.01  ch.  and  altitude  18.14  ch. 

5.  What  will  be  the  cost  of  a  triangular  piece  of  land  whose  base 
is  15.48  ch.  and  altitude  9.67  ch.,  at  $60  an  acre? 

6.  At  $.40  a  square  yard,  what  is  the  cost  of  paving  a  triangular 
court,  its  base  being  105  ft.  and  altitude  21  yards?     Ans.  $147. 

7.  Find  the  area  of  the  gable  end  of  a  house  that  is  28  ft,  wide, 
and  the  ridge  of  the  roof  15  ft.  higher  than  the  foot  of  the  rafters. 


424  MENSURATION. 

73O.  The  area  and  one  dimension  being  given  to  find 
the  other  dimension. 

1 .  What  is  the  base  of  a  triangle  whose  area  is  1 89  sq.  ft.  and 
altitude  14  ft.  ? 

OPERATION. — 189  sq.  ft.  x  2-5-14  =  27  ft.,  base, 

2.  Find  the  altitude  of  a  triangle  whose  area  is  20^  sq.  ft.  and 
base  3  yards.  Am.   4-J-  ft. 

RULE.     Double  the  area,  then  divide  by  the  given  dimension. 

Find  the  other  dimension  of  the  triangle 

3.  When  the  area  is  65  sq.  in.  and  the  altitude  10  in.  Ans.  13  i. 

4.  When  the  base  is  42  rd.  and  the  area  588  sq.  rd. 

5.  When  the  area  is  6 £  acres  and  altitude  1 7  yards. 

6.  When  the  base  is  12.25  ch.  and  the  area  5  A.  33  P. 

7.  Paid  $1050  for  a  piece  of  land  in  the  form  of  a  triangle,  at 
the  rate  of  $5^  per  square  rod.     If  the  base  is  8  rd.,  what  is  its 
altitude  ?  Ans.  50  rods. 


731.   The  three  sides  of  a  triangle  being  given  to  find  its 
area. 

1.  Find  the  area  of  a  triangle  whose  sides  are  30,  40,  and  50  ft. 

OPERATION.—  (30  +  40  +  50)  +2  =  60;  60-30=30;  60-40=20;  60-50 
=  10.       /§0>x30xlO  =  600  ft.,  area. 


2.  What  is  the  area  of  an  isosceles  triangle  whose  base  is  20  ft., 
each  of  its  equal  sides  15  ft.  ?  Ans.  111.85  sq.  ft. 

RULE,  from  half  the  sum  of  the  three  sides  subtract  each  side 
separately  ;  multiply  the  half-sum  and  the  three  remainders  together  ; 
the  square  root  of  the  product  is  the  area. 

3.  How  many  acres  in  a  field  in  the  form  of  an  equilateral  tri- 
angle whose  sides  measure  70  rods?  Ans.  13  A.  41.76  P. 

4.  The  roof  of  a  house  30  ft.  wide  has  the  rafters  on  one  side 
20  ft.  long,  and  on  the  other  18  ft.  long.     How  many  square  feet 
of  boards  will  be  required  to  board  up  both  gable  ends  ? 


TKIANGLES.  425 

The  following  principles  relating  to  right-angled  triangles 
have  been  established  Iby  Geometry : 

1.  The  square  of  the  hypothenuse  of 
a  right-angled  triangle  is  equal  to  the 
sum  of  the  squares  of  the  other  two 
sides. 

2.  The  square  of  the  base,  or  of  the 
perpendicular,    of  a   right-angled    tri- 
angle  is   equal   to    the   square   of  the 
hypothenuse  diminished  by  the  square 
of  the  other  side. 

733.  To  find  the  hypothemise, 

1.  The  base  of  a  right-angled  triangle  is  12,  and  the  perpendicu- 
lar 16.     What  is  the  length  of  the  hypothennse? 

OPERATION.— 122+162  =  400  (Prin.  1).     /y/400  =  20,  hypothenuse. 

2.  The  foot  of  a  ladder  is  15  ft.  from  the  base  of  a  building,  and 
the  top  reaches  a  window  36  ft.  above  the  base.    What  is  the  length 
of  the  ladder  ?  Ans.  39  ft. 

RULE.     Extract  the  square  root  of  the  sum  of  the  squares  of  the 
base  and  the  perpendicular  ;  and  the  result  is  the  hypothenuse. 

3.  If  the  gable  end  of  a  house  40  feet  wide  is  16  ft.  high,  what 
is  the  length  of  the  rafters  ? 

4.  A  park  25  ch.  long  and  23  ch.  wide  has  a  walk  running  through 
it  from  opposite  corners  in  a  straight  line.     What  is  the  length  of 
the  walk?  Ans.  33.97  ch.  -f . 

5.  A  room  is  20  ft.  long,  16  ft.  wide,  and  12  ft  high.     What  is 
the  distance  from  one  of  the  lower  corners  to  the  opposite  upper 
corner?  Ans.  28  ft.  3.36  in. 


734L  To  find  the  base  or  perpendicular. 

1.  The  hypothenuse  of  a  right-angled  triangle 
srpendicular  28  ft.      What  is  the  base  ? 

OPERATION.—  352  -  282  =  441  (Prin.  2).    ^441  =  21,  base. 


1.  The  hypothenuse  of  a  right-angled  triangle  is  35  ft.,  and  the 
perpendicular  28  ft.     What  is  the  base? 


426 


MENSURATIOX. 


2.  The  hypothenuse  of  a  right-angled  triangle  is  53  yd.  and  the 
base  84  feet.     Find  the  perpendicular. 

RULE.  Extract  the  square  root  of  the  difference  between  the  square 
of  the  hypothenuse  and  the  square  of  the  given  side  y  and  the  result  is 
the  required  side. 

3.  A  line  reaching  from  the  top  of  a  precipice  120  ft.  high,  on 
the  bank  of  a  river,  to  the  opposite  side  is  380  ft.  long.     How  wide 
is  the  river  ?  Ans.  360  ft.  6  in.  -j- . 

4.  A  ladder  52  feet  long  stands  against  the  side  of  a  building. 
How  many  feet  must  it  be  drawn  out  at  the  bottom  that  the  top 
may  be  lowered  4  feet?  Ans.  20  ft. 


QUADRILATERALS. 

735.  A  Quadrilateral  is  a  plane  figure  bounded  by  four  straight 
lines,  and  having  four  angles. 

There  are  three  kinds  of  quadrilaterals,  the  Parallelogram,  Trapezoid 
and  Trapezium. 

736.  A  Parallelogram  is  a  quadrilateral  which  has  its  oppo- 
site sides  parallel. 

There  are  four  kinds  of  parallelograms,  the  Square,  Rectangle,  Rhom- 
boid, and  Rhombus. 

737.  A  Eectangle  is  any  parallelogram  having  its  angles  right 
angles. 

738.  A  Square  is  a  rectangle  whose  sides  are  equal. 

73O.  A  Rhomboid  is  a  parallelogram  whose  opposite  sides  only 
are  equal,  but  whose  angles  are  not  right  angles. 

74O.  A  Rhombus  is  a  parallelogram  whose  sides  are  all  equal, 
but  whose  angles  are  not  right  angles. 


Square. 


Rectangle. 


Rhomboid. 


Rhombus. 


QUADRILATERALS. 


427 


7  4 1  o  A  Trapezoid  is  a  quadrilateral,  two  of  whose  sides  are 
parallel  and  two  oblique. 

743.  A  Trapezium  is  a  quadrilateral  having  no  two  sides 
parallel. 

743.  The  Altitude  of  a  parallelogram  or  trapezoid  is  the  per- 
pendicular distance  between  its  parallel  sides. 

The  vertical  dotted  lines  in  the  figures  represent  the  altitude. 

744.  A  Diagonal  of  a  plane  figure  is  a  straight  line  joining 
the  vertices  of  two  angles  not  adjacent. 


Parallelogram. 


Trapezoid. 


Trapezium. 


PROBLEMS. 

745.  To  find  the  area  of  any  parallelogram, 

1.  Find  the  area  of  a  parallelogram  whose  base  is  16.25  ft.  and 
altitude  7.5  feet. 

OPERATION.— 16.25  x  7.5  =  121.875  sq.  ft.,  area. 

2.  The  base  of  a  rhombus  is  10  ft.  6  in.,  and  its  altitude  8  ft. 
What  is  its  area  ? 

RULE.     Multiply  the  base  by  the  altitude. 

3.  How  many  acres  in  a  piece  of  land  in  the  form  of  athomboid, 
the  base  being  8.75  ch.  and  altitude  6  ch.  ?  Ans.  54-  A. 

746.  To  find  the  area  of  a  trapezoid. 

1.  Find  the  area  of  a  trapezoid  whose  parallel  sides  are  23  and 
11  feet,  and  the  altitude  9  feet. 

OPERATION.— 23  ft. +  11  ft.A-2  =  17  ft.  ;  17  ft.  x  9  —  153  sq.  ft.,  area. 

2.  Required  the  area  of  a  trapezoid  whose  parallel  sides  are  178 
and  146  feet,  and  the  altitude  69  feet.         •     Ans.   11178  sq.  ft. 

RULE.     Multiply  one-half  the  sum  of  the  parallel  sides  by  the 
altitude. 


428  MENSUKATION. 

3.  How  many  square  feet  in  a  board  16  ft.  long,  18  in.  wide  at 
one  end  and  25  in.  wide  at  the  other  end? 

4.  One  side  of  a  quadrilateral  field  measures  38  rd. ;  the  side 
opposite  and  parallel  to  it  measures  26  rd.,  and  the  distance  between 
the  two  sides  is  10  rd.     Find  the  area.  Ans.  2  A. 

74:7.  To  find  the  area  of  a  trapezium. 

1.  Find  the  area  of  a  trapezium  whose 
diagonal  is  42  ft.  and  perpendiculars  to  this 
diagonal,  as  in  the  diagram,  are  16  ft.  and  1 8  ft. 


OPERATION.— (18  it.  + 16  ft.  -r-2)  x  42  =  714  sq.  ft.,  area. 

2.  Find  the  area  of  a  trapezium  whose  diagonal  is  35  ft.  6  in.,  and 
the  perpendiculars  to  this  diagonal  9  ft.  and  12  ft.  6  in. 

RULE.  Multiply  the  diagonal  by  half  the  sum  of  the  perpendicu- 
lars drawn  to  it  from  the  vertices  of  the  opposite  angles. 

3.  How  many  acres  in  a  quadrilateral  field  whose  diagonal  is 
80  rd.  and  the  perpendiculars  to  this  diagonal  20.453  and  50.832  rd.? 

To  find  the  area  of  any  regular  polygon,  multiply  its  perimeter,  or  the 
sum  of  its  sides,  by  the  perpendicular  falling  from  its  centre  to  one  of 
its  sides. 

To  find  the  area  of  an  irregular  polygon,  divide  the  figure  into  triangles 
and  trapeziums,  and  find  the  area  of  each  separately.  The  sum  of  these 
areas  will  be  the  area  of  the  whole  polygon. 

THE     CIRCLE. 

748.  A  Circle  is  a  plane  figure  bounded  by  a  curved  line, 
called  the  circumference,  every  point  of  which  is 
equally  distant  from  a  point  within  called   the 
center. 


749.  The  Diameter  of  a  circle  is  a  line  pass- 
ing through  its  center,  and  terminated  at  both  ends 
by  the  circumference. 

7«5O.  The  Radius  of  a  circle  is  a  line  extending  from  its  center 
to  any  point  in  the  circumference.     It  is  one-half  the  diameter. 


CISCLE.  429 

PROBLEMS. 

.  When  either  the  diameter  or  the  circumference  of 
a  circle  is  given,  to  find  the  other  dimension  of  it. 

1.  Find  the  circumference  of  a  circle  whose  diameter  is  20  in. 
OPERATION.—  20  in.  x  3.1416  =  62.832  in.  =  5  ft.  2.832  in.,  circum. 

2.  What  is  the  circumference  of  a  wheel  5  ft.  6  in.  in  diameter. 

3.  Find  the  diameter  of  a  circle  whose  circumference  is  62.832  ft. 
OPERATION.—  62.832  ft.  -4-3.1416  =  20  ft.,  diameter. 

4.  Find  the  diameter  of  a  wheel  whose  circumference  is  50  ft. 
RULE  1.  Multiply  the  diameter  by  3.1416  ;  the  product  is  the  cir- 

cumference. 

2.  Divide    the   circumference    by    3.1416;     the   quotient   is    the 
diameter. 

5.  What  is  the  diameter  of  a  tree  whose  girt  is  18  ft.  6  in.? 

6.  Find  the  length  of  tire  that  will  band  a  wheel  7  ft.  9  in.  in 
diameter.  Ans.  24  ft.  4  in.  + 

7.  The  diameter  of  a  cylinder  is  8  ft.  6  in.  ;  find  its  girt. 

8.  What  is  the  radius  of  a  circle  whose  circumference  is  31.41  6  ft.  ? 

9.  The  radius  of  a  circle  is  10  ft.  ;  what  is  its  circumference? 

10.  Find  the  circumference  of  the  greatest  circle  that  can  be 
drawn  with  a  string  14  in.  long,  used  as  a  radius.       7  ft.  3.96  in. 


.  To  find  the  area  of  a  circle,  when  both  its  diam- 
eter and  circumference  are  given,  or  when  either  is  given. 

1.  Find  the  area  of  a  circle  whose  diameter  is  10  ft.  and  circum- 
ference 31.416  feet. 

OPERATION.—  31.410  ft.  xlO-s-4  =  78.54  sq.  ft.,  area. 

2.  Find  the  area  of  a  circle  whose  diameter  is  10  ft. 
OPERATION.—  10  ft.2  x  .7854  =  78.54  sq.  ft.,  area. 

3.  Find  the  area  of  a  circle  whose  circumference  is  31.416  ft. 
OPERATION.—  31.416  ft.  -5-  3.  1416  =  10  ft.,  diem.;  (10  ft.)2  x  .7854  = 

78.54  sq.  ft.  ,  area. 

RULES.     To  find  the  area  of  a  circle  : 

1.  Multiply  the  square  of  its  diameter  by  .7854. 

2.  Multiply  J  of  its  diameter  by  the  circumference. 


430  MENSURATION. 

4.  What  is  the  area  of  a  circular  pond  whose  circumference  is 
200  chains?  Ans.   318.3  A. 

5.  The  distance  around  a  circular  park  is  1 J  miles.     How  many 
acres  does  it  contain  ?  Ans.  114.59"  A. 

6.  Find  the  area  of  the  largest  circle  that  can  be  drawn  by  using 
as  a  radius  a  striiig  20  in.  long. 

753.  To  find  the  diameter  or  circumference  of  a  circle, 
when  the  area  is  given. 

1.  What  is  the  diameter  of  a  circle  whose  area  is  1319.472  ? 


OPERATION.— 1319. 472-J-.7854  =  1680  ;  V1680  =  40.987  +  ,  diameter. 
2.  What  is  the  circumference  of  a  circle  whose  area  is  19.635  ? 

OPERATION— 19.635  -f-  3.1416  =  6.25  ;    ^(^25=2.5,  radius;   2.5  x  2  x 
8.1416  =  15.708,  circumference. 

RULE  1.    Divide  the  area  by  .7854  and  extract  the  square  root  of 
the  quotient ;  the  result  is  the  diameter. 

2.  Divide  the  area  by  3.1416  and  extract  the  square  root  of  the 
quotient ;  the  result  is  the  radius.     The  circumference  is  obtained 
by  Art.  (751,  1).     Or, 

3.  Divide  the  area  by  .07958,  and  extract  the  square  root  of  the 
quotient. 

3.  The  area  of  a  circular  lot  is  38.4846  square  rods.     What  is 
its  diameter?  Ans.  7  rods. 

4.  The  area  of  a  circle   is  286.488  square  feet.     Required  the 
diameter  and  the  circumference. 

754.  To  find  the  side  of  an  inscribed  square  when  the 
diameter  of  the  circle  is  known. 

1.  What  is  the  side  of  a  square  inscribed  in  a 
circle  whose  diameter  is  6  rods? 

OPERATION.— 62  -s-  2  =  18  ;  \/lS  =  4.24  rd.,  side  of 
square. 

2.  The  diameter  of  a  circle  is  200  feet.     Find 
the  side  of  the  inscribed  square. 


CIRCULAR   EING.  431 

RULE.     Extract  the  square  root  of  half  the  square  of  the  diam- 
eter.    Or, 

Multiply  the  diameter  by  .7071. 

3.  The  circumference  of  a  circle  is  104  yards.     Find  the  side  of 
the  inscribed  square.  Ans.  23.4  yd.  +  . 

755.  To  find  the  area  of  a  circular  ring,  formed  by  two 
concentric  circles. 

1.  Find  the  area  of  a  circular  ring,  when  the 
diameters  of  the  circles  are  20  and  30  feet. 


OPERATION.—  (30  -f-  20  x  30  —  20)  x  .7854  =  392.7 
sq.  ft.,  area. 

2.  Find  the  area  of  a  circular  ring  formed  by 

two  concentric  circles,  whose  diameters  are  7  ft.  9  in.  and  4  ft.  3  in. 

Ans.  32.9868  sq.  ft, 

RULE.  —  Multiply  the  sum  of  the  two  diameters  by  their  difference, 
and  the  product  by  .7854;  the  result  is  the  area. 

3.  Two  diameters  are  35.75  and  16.25  feet  ;  find  the  area  of  the 
ring.  Ans.   796.39  sq.  ft. 

4.  The  area  of  a  circle  is  1  A.  154.16  P.     In  the  center  is  a  pond 
of  water  10  rods  in  diameter;  find  the  area  of  the  land  and  of  the 
water.  Ans.  Land,  235.62  P.  ;  water,  78.54  P. 

756.  To  find  a  mean  proportional  between  two  numbers. 
1.  What  is  a  mean  proportional  between  3  and  12  ? 
OPEKATION.  —  -\/12x3  =  6,  the  mean  proportional. 

When  three  numbers  are  proportional,  the  product  of  the  extremes  is 
equal  to  the  square  of  the  mean. 

RULE.     Extract  the  square  root  of  the  product  of  the  two  numbers. 
Find  a  mean  proportional  between 


2.  36  and     81. 

3.  42  and  168. 


4.  64  and    12.25. 

5.  8  and  288. 


6.  ff  and 

7.  f£  and 


8.  A  tub  of  butter  weighed  36  Ib.  by  the  grocer's  scales;  but 
being  placed  in  the  other  scale  of  the  balance,  it  weighed  only  30  Ib. 
What  was  the  true  weight  of  the  butter?  Ans.  32.86+  Ib. 


432  MENSURATIOK. 

SIMILAR     PLANE     FIGURES. 

757.  Similar  Plane  Figures  arc  such  as  have  the  same  form; 
or  have  angles  equal  each  to  each,  the  same  number  of  sides,  and 
the  sides  containing  the  equal  angles  proportional. 

All  circles,  squares,  equiangular  triangles,  and  regular  polygons  of  the 
same  number  of  sides  are  similar  figures. 

The  like  dimensions  of  circles,  that  is,  their  radii,  diameters,  and  cir- 
cumferences, are  proportional. 

PRINCIPLES.  1.  The  like  dimensions  of  similar  plane  figures  are 
proportional. 

2.  The  areas  of  similar  plane  figures  are  to  each  other  as  the 
squares  of  their  like  dimensions.     And  conversely, 

3.  The  like  dimensions  of  similar  plane  figures  are  to  each  other 
as  the  square  root  of  their  areas. 

The  same  principles  apply  also  to  the  surfaces  of  all  similar  figures, 
such  as  triangles,  rectangles,  etc.  ;  the  surfaces  of  similar  solids,  as  cubes, 
pyramids,  etc.  ;  and  to  similar  curved  surfaces,  as  of  cylinders,  cones, 
and  spheres.  Hence, 

4.  The  surfaces  of  all  similar  figures  are  to  each  other  as  the 
squares  of  their  like  dimensions.     And  conversely, 

5.  Their  dimensions  are  as  the  square  roots  of  their  surfaces. 

PROBLEMS. 

1.  A  triangular  field  whose  base  is  12  ch.  contains  2  A.  80  P. 
Find  the  area  of  a  field  of  similar  form  whose  base  is  48  ch. 

OPERATION.—  122  :  48*  :  :  2  A.  80  P.  :  x  =  6400  P  =  40  A.,  area. 

(PRIN.  2.) 

2.  The  side  of  a  square  field  containing  18  A.  is  60  rd.  long. 
Find  the  side  of  a  similar  field  that  contains  -J-  as  many  acres. 


OPERATION.—  18  A.  :  6  A.  :  :  602  :  tf  =  1200  ;  V^OO  =  34.64  rd.  +  ,  side. 
(PRIN.  3.) 

3.  Two  circles  are  to  each  other  as  9  to  16;  the  diameter  of  the 
less  being  112  feet,  what  is  the  diameter  of  the  greater? 

OPERATION.—  9  :  16  :  :  1122  :  x*=3  :  4  :  :  112  :  03=149  ft.  4  in.,  diameter. 
(PRIN.  2.) 

4.  A  peach  orchard  contains  720  sq.  rd.,  and  its  length  is  to  its 
breadth  as  5  to  4  ;  what  are  its  dimensions? 

OPERATION.—  The  area  of  a  rectangle  5  by  4  equals  20  (745). 
20  :  720  :  :  5s  :  tf  -  900  ;     ^900  =  30  rd.,  length. 
20  :  720  :  :  49  :  z5  =  576  ;     ^576  =  24  rd.,  width. 


REVIEW.  433 

5.  It  is  required  to  lay  out  283  A.  107  P.  of  land  in  the  form  of 
a  rectangle,  so  that  the  length  shall  be  3  times  the  width.     Find  the 
dimensions.  Ans.  369  rd. ;  123rd. 

6.  A  pipe  1.5  in.  in  diameter  fills  a  cistern  in  5  hr. ;  find  the  diam- 
eter of  a  pipe  that  will  fill  the  same  cistern  in  55  miu.  6  sec. 

Ans.  3.5  in. 

7.  The  area  of  a  triangle  is  24276  sq.  ft.,  and  its  sides  in  propor- 
tion to  the  numbers  13,  14,  and  15.     Find  the  length  of  its  sides 
ia  feet.  Ans.   221,  238,  and  256  ft. 

8.  A  field  containing  6  A.  is  laid  down  on  a  plan  to  a  scale  of 

1  in.  to  20  ft.     How  much  paper  will  it  cover?     Ans.  653. 4  sq.  in. 

9.  If  it  cost  $167.70  to  enclose  a  circular  pond  containing  17  A. 
110  P.,  what  will  it  cost  to  enclose  another  ^  as  large  ?    Ans.  $75. 

10.  If  a  cistern  6  ft.  in  diameter  holds  80  bbl.  of  water,  what  is 
the  diameter  of  a  cistern  of  the  same  depth,  that  holds  1 200  bbl. 

11.  If  63.39  rd.  of  fence  wilt  enclose  a  circular  field  containing 

2  A.,  what  length  will  enclose  8  A.  in  circular  form  ?  Ans.  126.78  rd. 

758.  REVIEW  OF  PLANE  FIGURES. 

PROBLEMS. 

1.  The  area  of  a  triangle  is  270  yd.,  and  the  perpendicular  45  ft. 
Find  the  base. 

2.  Find  the  area  of  a  square  whose  perimeter  is  the  same  as  that 
of  a  rectangle  48  ft.  by  28  feet. 

3.  A  rectangle  whose  length  is  3  times  its  width  contains  1323  P. 
Find  its  dimensions.  Ans.  21  rd.  by  63  rd. 

4.  Find  the  area  of  an  equilateral  triangle  whose  sides  are  36  ft. 

5.  The  area  of  a  circle  is  7569  square  feet.     Find  the  length 
of  the  side  of  a  square  of  equal  area.  Ans.  87  ft. 

6.  How  much  less  will  the  fencing  of  20  A.  cost  in  the  square 
form  than  in  the  form  of  a  rectangle  whose  breadth  is  -J-  the  length, 
the  price  being  $2.40  per  rod?  Ans.  $185.43. 

7.  A  house  that  is  50  ft.  long  and  40  ft.  wide  has  a  square  or 
pyramidal  roof,  whose  height  is  15  ft.     Find  the  length  of  a  rafter 
reaching  from  a  corner  of  the  building  to  the  vertex  of  the  roof. 

8.  Find  the  length  of  a  rafter  reaching  from  the  middle  of  ono 
side.  Ans.  25  ft. 


434  MENSUEATION. 

9.  Find  the  length  of  a  rafter  reaching  from  the  middle  of  one  end. 

10.  What  is  the  diameter  of  a  circular  island  containing  1J  sq. 
miles  ?  Ans.  403.7  rd. 

11.  How  many  rods  more  of  fencing  are  required  to  enclose  a 
square  field  whose  area  is  5  acres,  than  to  enclose  a  circular  field 
having  the  same  area? 

12.  What  is  the  value  of  a  farm,  at  $75  an  acre,  its  form  being  a 
quadrilateral,  with  two  of  its  opposite  sides  parallel,  one  40  chains 
and  the  other  22  chains  long,  and  the  perpendicular  distance  between 
them  25  chains  ?  Ans.  $5812.50. 

13.  What  is  the  difference  in  the  area  of  a  grass  plat  20  feet 
square  and  a  circular  plat  20  feet  in  diameter? 

14.  Find  the  cost,  at  18  cents  a  square  foot,  of  paving  a  space 
in  the  form  of  a  rhombus,  the  sides  of  which  are  15  ft.,  and  a  per- 
pendicular drawn  from  one  oblique  angle  will  meet  the  opposite 
side  9  feet  from  the  adjacent  angle.  Ans.  $32.40. 

1 5.  A  goat  is  fastened  to  the  top  of  a  post  4  feet  high  by  a  rope 
50  ft.  long.     Find  the  circumference  and  area  of  the  greatest  circle 
over  which  he  can  graze. 

16.  How  much  larger  is  a  square  circumscribing  a  circle  40  rd. 
in  diameter,  than  a  square  inscribed  in  the  same  circle? 

17.  What  is  the  value  of  a  piece  of  land  in  the  form  of  a  tri- 
angle, whose  sides  are  40,  48,  and  54  rods,  respectively,  at  the  rate 
of  $125  an  acre?  Ans.  $724.75. 

18.  The  radius  of  a  circle  is  5  feet;  find  the  diameter  of  anothei 
circle  containing  4  times  the  area  of  the  first.  Ans.  20  ft. 

19.  Find  the  difference  in  the  area  of  a  circle  36  feet  in  diameter, 
and  the  inscribed  square. 

20.  How  many  acres  in  a  semi-circular  farm,  whose  radius  i? 
100  rods?  Ans.   98  A.  28  P. 

21.  What  must  be  the  width  of  a  walk  extending  around  a  gar- 
den 100  feet  square,  to  occupy  one-half  the  ground? 

22.  An  irregular  piece   of  land,  containing  540  A.  36  P.,  is  ex- 
changed for  a  square  piece  of  the  same  area ;  find  the  length  of 
one  of  its  sides.     If  divided  into   42   equal  squares,  what  is  the 
length  of  the  side  of  each?  Ans.  to  last,  45.86  rd. 


SOLIDS. 


435 


23.  A  field  containing  15  A.  is  30  rd.  wide,  and  is  a  plane  inclining 
in  the  direction  of  its  length,  one  end  being  120  ft.  higher  than  the 
other.     Find  how  many  acres  of  horizontal  surface  it  contains. 

24.  If  a  pipe  3  inches  in  diameter  discharge  12  hogsheads  of 
water  in  a  certain  time,  what  must  be  the  diameter  of  a  pipe  which 
will  discharge  48  hogsheads  in  the  same  time?  Ans.  6  in. 


SOLIDS. 


759.  A  Solid  or  Body  has  three  dimensions,  length,  breadth, 
and  thickness. 
The  planes  which  bound  it  are  called  its  faces,  and  their  intersections, 


76®.  A  Prism  is  a  solid  whose  ends  are  equal  and  parallel 
polygons,  and  its  sides  parallelograms. 

Prisms  take  their  names  from  the  forms  of  their  bases,  as  triangular, 
quadrangular,  pentagonal,  etc. 

761.  The  Altitude  of  a  prism  is  the 
perpendicular  distance  between  its  bases. 

762.  A  Parallelopipedon  is  a  prism 
bounded  by  six  parallelograms,  the  opposite 
ones  being  parallel  and  cquai. 

763.  A    Cube    is   a   parallelopipedon 

whose  faces  are  all  equal  squares.  Cube> 

764.  A  Cylinder  is  a  body  bounded  by  a  uniformly  curved 
surface,  its  ends  being  equal  and  parallel  circles. 

1.  A  cylinder  is  conceived  to  be  generated  by  the  revolution  of  a  rectan- 
gle about  one  of  its  sides  as  an  axis. 

2.  The  line  joining  the  centres  of  the  bases,  or  ends,  of  the  cylinder  is 
its  altitude,  or  axis. 


Triangular 
Prisin. 


Quadrangular 
Pi-ism' 


Pentagonal 
Prism. 


Cylinder. 


436 


MENSUBATIOtf. 


PROBLEMS. 

765.  To  find  the  convex  surface  of  a  prism  or  cylinder. 

1.  Find  the  area  of  the  convex  surface 
of  a  prism  whose  altitude  is  7  feet,  and  its 
base   a   pentagon,   each  side   of   which  is 
4  feet, 

OPERATION. — 4  ft.  x  5  =  20  ft.,  perimeter. 
20  ft.  x  7  =  140  sq.  ft.,  convex  surface. 

2.  Find  the  area  of  the  convex  surface 
of  a  triangular  prism,  whose  altitude  is  8-J- 
feet,  and  the  sides   of   its  base  4,  5,  and 
6  feet,  respectively. 

OPERATION.  —4  ft.  +  5  ft.  +  6  ft.  =  15  ft.,  perim. 
15  ft.  x  8£  =  127£  sq.  ft.,  convex  surface. 

3.  Find  the  area  of  the  convex  surface  of  a  cylinder  whose  alti- 
tude is  2  ft.  5  in.  and  the  circumference 

of  its  base  4  ft.  9  in. 

OPERATION.— 2  ft.  5  in. =29  in. ;  4  ft.  9  in. 
=  57  in. 

57  in.  x  29  =  1653  sq.  in. =11  sq.  ft.  69  sq. 
inches,  convex  surface. 

RULE.     Multiply  the  perimeter  of  the  base  by  the  altitude. 
To  find  the  entire  surface,  add  the  area  of  the  bases  or  ends. 

4.  If  a  gate  8  ft.  high  and  6  ft.  wide  revolve  upon  a  point  in  its 
centre,  what  is  the  entire  surface  of  the  cylinder  described  by  it  ? 

5.  Find  the  superficial  contents,  or  entire  surface  of  a  parallelo- 
pipedon  8  ft.  9  in.  long,  4  ft,  8  in.  wide,  and  3  ft,  3  in.  high. 

6.  What  is  the  entire  surface  of  a  cylinder  formed  by  the  revo- 
lution about  one  of  its  sides  of  a  rectangle  that  is  6  ft.  6  in.  long 
and  4  fee    wide?  Ans.  263.89  sq.  ft. 

7.  Find  the  entire  surface  of  a  prism  whose  base  is  an  equilateral 
triangle,  the  perimeter  being  18  ft.,  and  the  altitude  15  feet? 


PYRAMIDS    AKD    COXES.  437 

766.  To  find  the  volume  of  any  prism  or  cylinder. 

1.  Find  the  volume  of  a  triangular  prism,  whose  altitude  is  20  ft., 
and  each  side  of  the  base  4  feet. 
OPERATION.— The  area  of  the  base  is  6.928  sq.  ft.  (729). 
6.928  sq.  ft.  x  20  =  138.58  cu.  ft.,  volume. 

'2.  Find  the  volume  of  a  cylinder  whose  altitude  is  8  ft.  G  in., 
and  the  diameter  of  its  base  3  feet. 
OPERATION.— 32  x  .7854  =  7.0686  sq.  ft.,  area  of  base  (752). 
7.0686  sq.  ft.  x  8.5  —  60.083  cu.  ft.,  volume. 
RULE.     Multiply  the  area  of  the  base  by  the  altitude. 

3.  What  is  the  volume  of  a  parallelopipedon,  whose  base  is  9.8  ft. 
by  7.5  ft.,  and  its  height  5  ft.  3  in. 

4.  What  is  the  volume  of  a  log  18  ft.  long  and  1J-  ft.  in  diameter? 

5.  Find  the  solid  contents  of  a  cube  whose  edges  are  6  ft.  6  in.? 

6.  Find  the  cost  of  a  piece  of  timber  18  in.  square  and  40  ft. 
long,  at  $.30  a  cubic  foot.  Ans.  $27. 

7.  Required  the  solid  contents  of  a  cylinder  \vhosc  altitude  is 
15  ft.  and  its  radius  1  ft.  3  in.  Ans.  73.63  cu.  ft. 

PYRAMIDS     AND     COXES. 

767.  A  Pyramid  is  a  body,  having  for  its  base  a  polygon,  and 
for  its  other  faces  three  or  more  triangles,  which  terminate  in  a  com- 
mon point  called  the  vertex. 

Pyramids,  like  prisms,  take  their  names  from  their  bases,  and  are 
called  triangular,  square,  or  quadrangular,  pentagonal,  etc. 


Pyramid.  Frustum.  Cone.  Frustum. 

^68 •>  A  Cone  is  a  body  having  a  circular  base,  and  whose  con- 
vex surface  tapers  uniformly  to  the  vertex. 

A  cone  is  a  body  conceived  to  be  formed  by  the  revolution  of  a  right- 
angled  triangle  about  one  of  its  sides  containing  the  right  angle. 

76O.  The  Altitude  of  a  pyramid  or  of  a  cone  is  the  perpendic- 
ular distance  from  its  vertex  to  the  plane  of  its  base. 


438  MENSURATION. 

770.  The  Slant  Height  of  a  pyramid  is  the  perpendicular  dis- 
tance from  its  vertex  to  one  of  the  sides  of  the  base;  of  a  cone,  is  a 
straight  line  from  the  vertex  to  the  circumference  of  the  base. 

771.  The  Frustum    of  a  pyramid  or  cone  is  that  part -which 
remains  after  cutting  off  the  top  by  a  plane  parallel  to  the  base. 

THE     SPHERE. 

772.  A  Sphere  is  a  body  bounded  by  a  uniformly  curved  sur- 
face, all  the  points  of  which  are  equally  distant 

from  a  point  within  called  the  center. 

778.  The  Diameter  of  a  sphere  is  a  straight 
line  passing  through  the  center  of  the  sphere, 
and  terminated  at  both  ends  by  its  surface. 

774.  The  Radius  of  a  sphere  is  a  straight  line  drawn  from  the 
center  to  any  point  in  the  surface. 

PROBLEMS. 

775,  To  find  the  convex  surface  of  a  pyramid  or  cone. 

1 .  Find  the  convex  surface  of  a  triangular  pyramid,  the  slant 
height  being  16  ft.,  and  each  side  of  the  base  5  feet. 

OPERATION.— (5  ft. +  5  ft.  +5  ft.)  x  16^2 =120  sq.  ft.,  convex  surface. 

2.  Find  the  convex  surface  of  a  cone  whose  diameter  is  17  ft. 
6  in.,  and  the  slant  height  30  feet. 


OPEKATION.— 17.5  ft.  x  3.1416=54.978  ft.,  circum. ;  54.978  ft.  x  30-=-2= 
824.67  sq.  ft.,  convex  surface. 

RULE. — Multiply  the  perimeter  or  circumference  of  the  base  by 
one-half  the  slant  height. 

To  find  the  entire  surface,  add  to  this  product  the  area  of  the  base. 

3.  Find  the  entire  surface  of  a  pyramid  whose  base  is  8  ft.  6  in. 
square,  and  its  slant  height  21  feet. 

4.  Find  the  entire  surface  of  a  cone  the  diameter  of  whose  base 
is  6  ft.  9  in.  and  the  slant  height  45  feet.  Ans.  512.9  sq.  ft. 

5.  Find  the  cost  of  painting  a  church  spire,  at  $.25  a  sq.  yd.,  whose 
base  is  a  hexagon  5  ft.  on  each  side,  and  the  slant  height  60  feet. 


PYRAMIDS    AND    COKES.  439 

77G.  To  find  the  volume  of  a  pyramid  or  of  a  cone. 

1.  What  is  the  volume,  or  solid  contents,  of  a  square  pyramid 
whose  base  is  6  feet  on  each  side,  and  its  altitude  12  feet. 

OPERATION. — 6  x  6  x  12  -f-  3  =  144  cu.  ft.,  volume. 

2.  Find  the  volume  of  a  cone,  the  diameter  of  whose  base  is  5  ft. 
and  its  altitude  10|-  feet. 

OPERATION.— 52   ft.  x  .7854 x  10JT3  =  68.72£  cu.  ft.,  volume. 
RULE. — Multiply  the  area  of  the  base  by  one-third  the  altitude. 

3.  Find  the  solid  contents  of  a  cone  whose  altitude  is  24  ft,  and 
the  diameter  of  its  base  30  inches. 

4.  What  is  the  cost  of  a  triangular  pyramid  of  marble,  whose 
altitude  is  9  feet,  each  side  of  the  base  being  3  feet,  at  $2^-  per 
cubic  foot?  Ans.  $29.25. 

5.  Find  the  volume  and  the  entire  surface  of  a  pyramid  whose 
base  is  a  rectangle  80  ft.  by  60  ft,  and  the  edges  which  meet  at  the 
vertex  are  130  feet,  Ans.   192000  cu.  ft.  voL 

777.  To  find  the  convex  surface  of  a  frustum  of  a  pyra- 
mid or  cone. 

1.  What  is  the  convex  surface  of  a  frustum  of  a  square  pyramid, 
whose  slant  height  is  7  ft.,  each  side  of  the  greater  base  4  ft.,  and 
of  the  less  base  1 8  inches  ? 

OPERATION. — The  perimeter  of  the  greater  base  is  16  ft.,  of  the  less  6  ft. 
16  ft. +  6  ft.  x7-r-2  =  77  sq.  ft.,  convex  surface. 

2.  Find  the  convex  surface  of  a  frustum  of  a  cone  whose  slant 
height  is  15  ft.,  the  circumference  of  the  lower  base  30  ft.,  and 
of  the  upper  base  16  feet. 

RULE. — Multiply  the  sum  of  the  perimeters,  or  circumferences,  by 
one-half  the  slant  height. 

To  find  the  entire  surface,  add  to  this  product  the  areas  of  both  ends, 
or  bases. 

3.  How  many  square  yards  in  the  convex  surface  of  a  frustum 
of  a  pyramid,  whose  bases  are  heptagons,  each  side  of  the  lower 
base  being  8  feet,  and  of  the  upper  base  4  feet,  and  the  slant  height 
55  feet?  Ans.  256-|  sq.  yd. 


440  MEKSUEATIOK". 

778.  To  find  the  volume  of  a  frustum  of  a  pyramid  or 
cone. 

1.  Find  the  volume  of  a  frustum  of  a  square  pyramid  whose  alti- 
tude is  10  feet,  each  side  of  the  lower  base   12  feet,  and  of  the 
upper  base  9  feet. 

OPERATION.— 122  +  92  =  225 ;  (225  +  ^144  x  81)  x  10 -T- 3  =  1000  cu.  ft., 

2.  How  many  cubic  feet  in  the  frustum  of  a  cone  whose  altitude 
is  6  feet  and  the  diameters  of  its  bases  4  feet  and  3  feet? 

RULE. — To  the  sum  of  the  areas  of  both  bases  add  the  square  root 
of  the  product,  and  multiply  this  sum  by  one-third  of  the  altitude. 

3.  How  many  cubic  feet  in  a  piece  of  timber  30  ft.  long,  the 
greater  end  being  15  in.  square,  and  that  of  the  less  12  in.  ? 

4.  How  many  cubic  feet  in  the  mast  of  a  ship,  its  height  being 
50  ft.,  the  circumference  at  one  end  5  ft.  and  at  the  other  3  feet. 

779.  To  find  the  surface  of  a  sphere. 

1.  Find  the  surface  of  a  sphere  whose  diameter  is  9  inches. 
OPERATION.— 9  in.  x  3.1416  =  28.2744  in.,  circumference. 

28.2744  in.  x  9  =  254.4696  sq.  in.,  surface. 

RULE. — Multiply  the  diameter  by  the  circumference  of  a  great  cir- 
cle of  the  sphere. 

2.  What  is  the  surface  of  a  globe  1 6  in.  in  diameter  ? 

3.  Find  the  area  of  the  surface  of  a  sphere  whose  circumference 
is  31.416  feet.  Ana.  314.16  sq.  ft. 

4.  Find  the  surface  of  a  globe  whose  radius  is  1  foot. 

780.  To  find  the  volume  of  a  sphere. 

1.  Find  the  volume  of  a  sphere  whose  diameter  is  18  inches. 
OPERATION. — 18  in.  x  3.1416  =  56.5488  in.,  circumference. 

56.5488  in.  x  18  =  1017.8784  sq.  in.,  surface. 
1017.8784  sq.  in.  x  18-f-6  =  3053.6352  cu.  in.,  volume. 
RULE. — Multiply  the  surface  by  ^  of  the  diameter,  or  ^  of  the 
radius. 

2.  Find  the  volume  of  a  globe  whose  diameter  is  30  in. 

3.  Find  the  solid  contents  of  a  globe  whose  radius  is  5  yards. 

4.  Find  the  volume  of  a  globe  whose  circumference  is  31.416  ft. 


EEVIEW    OF    SOLIDS.  441 

781.  To  find  the  three  dimensions  of  a  rectangular  solid, 
the  volume  and  the  ratio  of  the  dimensions  being  given. 

1.  What  are  the  dimensions  of  a  rectangular  solid,  whose  volume 
is  4480  cu,  ft.,  and  its  dimensions  are  to  each  other  as  2,  5,  and  7  ? 


OPERATION.— ^/4480-*-(2  x  6  x  7)  =  4  ft. ;  4  x  2  =  8  ft.,  height;  4x5  = 
20  ft.,  width  ;  4  x  7  =  28  ft.,  length. 

RULE. — I.  Divide  the  volume  by  the  product  of  the  terms  propor- 
tional to  the  three  dimensions,  and  extract  the  cube  root  of  the  quo- 
tient. 

II.  Multiply  the  root  thus  obtained  by  each  proportional  term  ; 
the  products  will  be  the  corresponding  sides. 

2.  What  are  the  dimensions  of  a  rectangular  box  whose  volume 
is  3000  cu.  ft.,  and  its  dimensions  are  to  each  other  as  2,  3,  and  4  ? 

3.  A  pile  of  bricks  in  the  form  of  a  parallelepiped  contains  30720 
cu.  ft.,  and  the  length,  breadth,  and  height  are  to  each  other  as  3, 
4,  and  5.     What  are  the  dimensions  of  the  pile  ? 

4.  Separate  405  into  three  factors,  which  shall  be  to  each  other 
as  2,  2J,  and  3.  Ann.  6,  7£,  and  9. 

SIMILAR     SOLIDS. 

782.  Similar  Solids  are  such  as  have  the  same/orw,  and  differ 
from  each  other  only  in  volume. 

PRINCIPLES. — 1.  The  volumes  of  similar  solids  are  to  each  other 
as  the  cubes  of  their  like  dimensions. 

If  the  volume  of  a  ball  3  in.  in  diameter  is  27  cii.  in.,  what  is  the 
volume  of  a  ball  7  in.  in  diameter? 
OPERATION. — 33 :  73 : :  27  cu.  in. :  x  =  343  cu.  in.,  volume. 

2.  The  like  dimensions  of  similar  solids  are  to  each  other  as  the 
tube  roots  of  their  volumes. 

If  the  diameter  of  a  ball  whose  volume  is  27  cu.  in.  is  3  in.,  what 
is  the  diameter  of  a  ball  whose  volume  is  343  cu.  in.? 

OPERATION.— 27  :  343  : :  3s  :  x*  =  373 ;  ^/373  =  7  in.,  diameter. 


44:2  MENSUBATION. 

783.  REVIEW  OF  SOLIDS. 

PROBLEMS. 

1.  What  is  the  edge  of  a  cube  whose  entire  surface  is  1050  sq.  ft., 
and  what  is  its  volume  ? 

2.  What  must  be  the  inner  edge  of  a  cubical  bin  to  hold  1250 
bushels  of  wheat?  Ans.  11  ft.  7  in. 

3.  How  many  globes  4  in.  in  diameter  are  equal  to  one  whose 
diameter  is  12  inches? 

4.  How  many  gallons  will  a  cistern  hold,  whose  depth  is  7  ft., 
the  bottom  being  a  circle  7  ft.  in  diameter  and  the  top  5  feet  in 
diameter?  Ans.  1494.25  gal. 

O 

5.  What  is  the  value  of  a  stick  of  timber  24  ft.  long,  the  larger 
end  being  15  in.  square,  and  the  less  6  in.,  at  28  cents  a  cubic  foot? 

6.  If  a  cubic  foot  of  iron  were  formed  into  a  bar  -J  an  inch  square, 
without  waste,  what  would  be  its  length  ?  Ans.   576  ft. 

7.  How  many  barrels  of  31-J  gal.  will  a  cistern  hold  that  is  8.3  ft. 
in  diameter,  and  7  feet  deep? 

8.  If  a  log  18  ft.  long  and  3  ft.  in  diameter  is  hewn  square,  how 
many  cubic  feet  does  it  contain  ? 

9.  Find  the  volume  of  a  cube,  the  area  of  whose  entire  surface  is 
7  sq.  ft.  6  sq.  inches.  Ans.   1  cu.  ft.  469  cu.  in. 

10.  If  a  marble  column   10  in.  in  diameter  contains  27  cu.  ft., 
what  is  the  diameter  of  a  column  of  equal  length  that  contains  81 
cubic  feet?  Ans.  14.42  in. 

11.  Supposing  the  earth  to  be  a  perfect  sphere  7912  miles  in 
diameter,  what  is  its  volume  in  cubic  miles  ? 

12.  How  many  board  feet  in  a  post  11  ft.  long,  9  in.  square  at 
the  bottom,  and  4«in.  square  at  the  top?  Ans.  40  ft.  7-f. 

13.  The  surface  of  a  sphere  is  the  same  as  that  of  a  cube,  the 
edge  of  which  is  12  in.     Find  the  volume  of  each. 

14.  The  contents  of  a  cubical  block  of  marble  are  4913  cu.  ft. 
Find  the  superficial  contents  or  surface.  Ans.  1014  sq.  ft. 

15.  Find  the  dimensions  of  a  bin  that  holds  450  bu.  of  grain,  if 
the  width  and  depth  are  equal,  and  the  length  3  times  the  width. 

16.  A  ball  4.5  in.  in  diameter  weighs  18  oz.  Avoir.;  what  is  the 
weight  of  a  ball  of  the  same  density,  that  is  9  in.  in  diameter? 


GAUGING.  443 

17.  In  what  time  will  a  pipe  supplying  6  gal.  of  water  a  minute, 
fill  a  tank  in  the  form  of  a  hemisphere,  that  is  10  ft.  in  diameter? 

18.  If  the  altitude  of  a  cone  that  weighs  640  Ib.  is  8  ft,,  what  is 
the  altitude  of  a  similar  cone  that  weighs  270  Ib.  ? 

19.  If  a  stack  of  hay  8  ft.  high  weighs  8  cwt.,  what  is  the  weight 
of  a  similar  stack  that  is  24  ft.  high?  Ans.  216  cwt. 

20.  The  diameter  of  a  cistern  is  8  ft. ;  what  must  be  its  depth 
to  contain  75  hhd.  of  water?  Ans.  12.56  ft. 

21.  If  a  cable  3  inches  in  circumference  supports  a  weight  of 
2500  pounds,  what  must  be  the  circumference  of  a  cable  that  will 
support  4960  pounds?  Ans.  3.77  in.,  nearly. 

22.  How  many  bushels  in  a  heap    of   grain  in  the  form  of  a 
cone,  whose  base  is  8  ft.  in  diameter  and  altitude  4  feet  ? 

GAUGING. 

784.  Gauging  is  the  process  of  finding  the  capacity  or  volume 
of  casks  and  other  vessels. 

For  ordinary  purposes  the  diagonal  rod  is  used, 
which  gives  only  approximate  results. 

A  cask  is  equivalent  to  a  cylinder  having  the 
same  length  and  a  diameter  equal  to  the  mean 
diameter  of  the  cask. 

785.  To  find  the  mean  diameter  of  a  cask  (nearly). 
RULE.    Add  to  the  head  diameter  -|,  or,  if  the  staves  are  but  little 

curved,  .6,  of  the  difference  between  the  head  and  bung  diameters. 

786.  To  find  the  volume  of  a  cask  in  gallons. 

RULE.     Multiply  the  square  of  the  mean  diameter  by  the  length 
(both  in  inches)  and  this  product  by  .0034. 

1.  How  many  gallons  in  a  cask  whose  head  diameter  is  24  in., 
bung  diameter  30  in.,  and  its  length  34  inches? 

OPERATION.— 24  +  (30  —  24  x  f )  =  28  in.,  mean  diameter. 
282  x  34  x  .0034  =  90.63  gal.,  capacity. 

2.  What  is  the  volume  of  a  cask  whose  length  is  40  in.,  the 
diameters  21  and  30  in.,  respectively?  Ans.  99.14  gal. 

3.  How  many  gallons  in  a  cask  of  slight  curvature,  3  ft.  6  in, 
long,  the  head  diameter  being  26  in.,  the  bung  diameter  31  in.  ? 


444  MENSTJKATIOH. 

787.  CIRCLES. 

1.  The  diameter  of  any  circle 

Multiplied),     j  3.1416,  the  product  )        ,,       . 

tv  M  a      f  &y  }     o-ioo    i  •       f  ==  the  circumference. 

Divided      )     J   (    .3 183,  the  quotient  j 

Multiplied  )  ,     j    .8862,  the  product  ) 

Divided      |  ^  1 1.1284,  the  quotient  f  =    **  «**  °fan  <*"* 
Multiplied  )  ^    j    .8660,  the  product  )  =  the  side  of  an  inscribed 
Divided      [    y  (    .1547,  the  quotient  j        equilateral  triangle. 
Multiplied  [  ^    j    .7070,  the  product  )  =  the  side  of  an  inscribed 
Divided      f  .    (  1.4142,  the  quotient  [       square. 

2.  The  radius  of  any  circle 

Multiplied  )  ,      (  6.28318) 

-n-  •  i  j       f  b7  )       ,  rrn-,  r  r  —  the  circumference. 

Divided      )     3  (    .15915  j 

3.  The  square  of  the  diameter  of  any  circle 
Multiplied  )  ,      (    .7854,  the  product  ) 

Divided      rn  1.2732,  the  quotient  |=    thc  "^ 

4.  The  circumference  of  any  circle 
Multiplied  )         (    .3183,  tlie  product  |  = 
Divided      j     J  (  3.1416,  the  quotient  \ 

Multiplied),      (    .2821,  the  product  ) 

T^  -j  j      r  by  1  0  r  ,.       h  =  theme  of  aw  equal sof re, 

Divided      )     "    (  3.5450,  the  quotient  j 

Multiplied)  .     j    .2756,  the  product  )  =  the  side  of  the  inscribed 
Divided     j        (  3.6276,  the  quotient  j        equilateral  triangle. 
Multiplied  )  .     j    .2251,  the  product  )  =  the  side  of  an  inscribed 
Divided      J     ^  (  4.4428,  the  quotient  j        square. 

5.  The  square  of  the  circumference  of  any  circle 
Multiplied  |        (      .07958,  the  product  )        ^ 
Divided      j     J  (  12.5663,    the  quotient  j 

6.  The  area  of  any  circle 

Multiplied  |        (  1.2732,  the  product  )  =  ^  re  o/  ^  rf.^ 

Divided     )     ^   (    .7854,  the  quotient  j 

(  The  square  of  the  radius  of  any  circle  x  3.1416        ^ 
7.   •<  Half  the  circumference  of  a  circle  x  \  its  diameter  >•  =  area. 
(  Square  of  the  circumference  of  a  circle  x  .07958     ) 


LAND.  445 

788.  SPHERES. 

r  Circumference  x  its  diameter. 
Radius2  x  12.5664. 

1.  The  Surface  H  Diameter*  x  3.1410. 

I  Circumference2  x  .3183. 

f  Surface  x  -J-  its  diameter. 

Radius3  x  4.1 888. 

2.  The  Volume  =  <  _.. 

I  Diameter3x.5236. 

I  Circumference3  x  .0169. 

3.  The  Diameter  =  j 


A/Of  surface  x  .5642. 
A/Of  volume  x  1.2407. 


(  A/Of  surface  x  1.77255. 

4.  The  Circumference  =  1  3  . 

(  A/Of  volume  x  3.8978. 


(  A/Of  surface  x  .2821. 
5.  The  Radius  =  -J   3  ,— — 

(  A/Of  volume  x  .6204. 

(  Radius  x  1.1 547. 
6.Thesideofamnscnbedcube=  4  DiametcrXe, 


LAND. 

?"8O.  The  Unit  of  land  measure  is  the  acre. 
Measurements  of  land  are  commonly  recorded  in  square  miles,  acres, 
and  hundredths  of  an  acre. 

PROBLEMS. 

1.  What  is  the  value  of  a  farm  189.5  rods  long  and  150  rods 
wide,  at  $42J  an  acre  ? 

2.  A  man  having  a  field  70  rd.  square  appropriated  5  A.  of  it  to 
corn,  100  sq.  rd.  to  garden  vegetables,  and  the  remainder  to  meadow. 
What  fractional  part  of  the  whole  field  did  the  meadow  comprise  ? 

3.  I  bought  a  piece  of  land  16  ch.  long  and  15  ch.  wide,  at  $100 
an  acre,  and  dividing  it  into  lots  of  6  rods  by  5  rods,  sold  them  at 
$50  each.     What  was  my  gain  ?  Ans.  $4000. 

4.  At  $2.75  a  rod,  how  much  less  will  it  cost  to  fence  a  piece  of 
land  80  rods  square,  than  if  the  same  were  in  the  form  of  a  rectangle 
twice  as  long  and  one-half  as  wide  ?  Ans.  $220. 


446 


MENSURATION. 


T9O.  Government  Lands  are  usually  surveyed  into  rectangular 
tracts,  bounded  by  lines  conforming  to  the  cardinal  points  of  the 
compass. 

A  Base-line  on  a  parallel  of  latitude,  and  a  Principal  Meridian 
intersecting  it,  are  first  established.  Other  lines  are  then  run  six 
miles  apart,  each  way,  as  nearly  as  possible. 

The  tracts  thus  formed  are  called  Townships,  and  contain,  as 
near  as  may  be,  23040  acres. 

A  line  of  townships  extending  north  and  south  is  called  a  Range* 

The  ranges  are  designated  by  their  number  east  or  west  of  the  princi- 
pal meridian. 

The  townships  in  each  range  are  designated  by  their  number  north  or 
south  of  the  base-line. 

Since  the  earth's  surface  is  convex,  the  principal  meridians  converge 
as  they  proceed  northward.  This  tends  to  throw  the  townships  and  sec- 
tions out  of  square,  and  necessitates  occasional  lines  of  offset,  called 
"  correction  lines." 

Townships  are  subdivided  into  Sections,  and  sections  into  Half- 
Sections,  Quarter- Sections,  Half -Quarter- Sections,  Quarter- Quarter- 
Sections,  and  Lots. 


Diagram  No.  1. 

A  TOWNSHIP. 

N 


Diagram  No.  2. 


6 

5 

4 

3 

2 

1 

7 

8 

9 

10 

11 

12 

18 

17 

16 

15 

14 

13 

19 

20 

21 

22 

23 

24 

30 

29 

28 

27 

26 

25 

31 

32 

33 

34= 

35 

36 

A  SECTION. 

N 

N.W.  ^ 

N.  W.  Y± 
4O  A. 

E,X 

of 
N.W.H 
80  A. 

N,  E,  H 
ISO  A. 

S.  W.  H 

of 

N.  W.  * 

8.* 

320  A. 

Diagram  No.  1  shows  the  sub-divisions  of  a  Township  into  Sections, 
and  how  they  are  numbered,  commencing  at  the  N.  E.  corner. 

Diagram  No.  2  shows  the  sub-divisions  of  a  Section,  on  an  enlarged 
icale,  and  how  they  are  named. 


LAND.  447 


TABLE. 

6  mi.  x  6  mi.  =  36  sq.  mi.  =  23040  Acres  =  1  Township. 

1  "  x  1  "    =    1      "       =      640      "      =1  Section. 

1   "  x  £  "    =    |      "       =       320      "      =  1  Half-Section. 

$  "  x  $  "    =    £      "       =      160      "      =  1  Quarter-Section. 

I  "  x  I  "    =    J      "       =        80      "      =1  Half -Quarter-Section. 

$  "  x  -} :  "    =  TV      "       =        40      "      —  1  Quarter-Quarter-Section, 

A  Lot  is  a  subdivision  of  a  section,  usually  of  irregular  form,  on  account 
of  bordering  upon  a  navigable  river  or  lake — containing  as  near  as  may 
be  the  area  of  a  Quarter-Quarter-Section,  and  described  as  lot  No.  1, 2,  3, 
etc.,  of  a  particular  section. 

City  and  village  plats  are  usually  sub-divided  into  Blocks,  and  these 
Into  Lots. 

PROBLEMS. 

1.  If  a  township  of  land  is  equally  divided  among  288  families, 
how  many  acres  does  each  receive  ?     What  part  of  a  section  ? 
.   2.   What  number  of  rails  will  enclose  a  quarter-section   of  land 
with  a  fence  6  rails  high,  and  3   lengths  for  every  2  rods ;    and 
what  will  be  the  cost  of  the  rails,  at  $40  per  thousand? 

3.  A  man  bought  the  S.  -|-  of  a  section  of  land  at  $2J  an  acre, 
and  afterward  sold  the  E.  -|  of  what  he  bought  at  $4.3  7J  an  acre. 
What  did  he  gain  on  what  he  sold?  Ans.  8340. 

4.  If  I  buy  the  N.  E.  J  and  the  E.  £  of  N.  W.  J-  of  a  section  of 
land,  how  many  acres  do  I  purchase  ?     What  part  of  a  whole  sec- 
tion ?     How  are  the  parts  located  in  respect  to  each  other  ? 

5.  A  speculator  bought  of  the  111.  Central  R.  R.  Co.,  the  S.  £  of 
Section  4,  township  10  north,  range  6  east,  at  $2  an  acre.     He  after- 
ward sold  the  E.  -J-  of  S.  E.  J  at  82.75  an  acre  ;  the  N.  E.  £  of  S.  E. 
J-  at  $3£  an  acre ;  and  the  N.  £  of  S.  W.  £  at  $3.84  an  acre.     How 
many  acres  has  he  left?     What  was  his  gain  on  the  purchase  price 
of  the  whole?     Draw  diagram.  Ans.  $27.20. 

6.  A  man  having  purchased  a  section  of  land  from  the  U.  S. 
Government  at  $1.25  an  acre,  sold  the  S.  \  of  S.  W.  J  at  $2.50  an 
acre;  the  N.W.  J  of  N.W.  J  at  81.75  an  acre;  the  W.  \  of  S.E.  } 
at  $2  an  acre;  and  the  W.  \  of  S.W.  J  of   N.E.  £  at  $3  an  acre. 
How  many  acres  has  he  remaining,  and  what  is  his  gain,  provided 
the  remainder  is  sold  at  82-J-  an  acre  ?     Draw  diagram  and  explain. 


448  MENSURATION. 


BOARDS     AND     TIMBER. 

791.  A  Board  Foot  is  I  ft.  long,  1  ft.  wide,  and  1  inch  thick. 
Hence  12  board  feet  make  1  cubic  foot. 

Board  feet  are  changed  to  cubic  feet  by  dividing  by  12,  and 
cubic  feet  are  changed  to  board  feet  by  multiplying  by  12. 

1.  In  Board  Measure  all  boards  are  assumed  to  be  1  in.  thick. 

2.  Lumber  and  sawed  timber,  as  plank,  scantling,  etc.,  are  usually  esti- 
mated in  board  measure,  hewn  and  round  timber  in  cubic  measure. 

When  lumber  is  not  more  than  1  inch  thick : 

RULES.  1.  Multiply  the  length  in  feet  by  the  width  in  inches,  and 
divide  the  product  by  12. 

When  more  than  1  inch  thick : 

2.  Multiply  the  length  infect  by  the  width  and  thickness  in  inches, 
and  divide  the  product  by  12. 

PROBLEMS. 

1.  What  must  be  the  width  of  a  board  16  ft.  long  to  contain  12 
board  feet  ? 

OPERATION.— 16  ft.  =  192  in.  ;  144  x  12  -f-  192  =  9  in.,  the  width. 

2.  What  must  be  the  width  of  a  piece  of  board  5  ft.  3  in.  long, 
to  contain  7  square  feet? 

3.  Find  the  cost  of  8  pieces  of  scantling,  3  in.  by  4  in.  and  14  ft. 
long,  at  $9.50  per  thousand  board  feet. 

4.  A  piece  of  timber  is  10  in.  by  16  in. ;  what  length  of  it  will 
contain  1 5  cubic  feet  ?  Ans.  1 3£  ft. 

5.  How  many  board  feet  in  a  stick  of  timber  36  ft.  long,  10  in. 
thick,  1 2  in.  wide  at  one  end  and  9  in.  wide  at  the  other  end  ? 
How  many  cubic  feet?  Ans.   26J  cu.  ft. 

6.  A  rectangular  field,  16  ch.  long  and  8  ch.  wide,  is  enclosed  by 
a  post  and  board  fence ;  the  posts  are  set  8  ft.  apart,  the  boards  are 
16  ft.  long,  and  the  fence  is  5  boards  high.     The  bottom  board  is 
12  in.  wide,  the  top  board  6  in.,  and  the  other  three  9  in.  wide, 
The  posts  cost  $25  per  C.,  and  the  boards  $14.80  per  M.     Required 
the  number  of  posts,  the  amount  of  lumber,  and  the  cost  of  both. 


MASONRY.  449 


MASONRY. 

792.  Masonry  is  estimated  by  the  cubic  foot,  and  by  the  perch  ; 
also  by  the  square  foot  and  the  square  yard  (287). 

1.  Brickwork  is  generally  estimated  by  the  thousand  bricks;  sometimes 
in  cubic  feet. 

2.  When  stone  is  built  into  a  wall,  22  cubic  feet  make  a  perch,  2f  cubic 
feet  being  allowed  for  mortar  and  filling. 

3.  Philadelphia  bricks  are  8|  in.  x  4£-  x  2f ;  and  Milwaukee  bricks 
8Hn.x4£-x2*. 

793.  To  find  the  number  of  bricks  in  a  cu.  ft.  of  masonry. 

PROBLEMS. 

1.  How  many  Milwaukee  bricks  in  a  cubic  foot  of  wall  12|  in. 
wide,  laid  in  courses  of  inortar  £  of  an  inch  thick  ? 

OPEKATION. 

8.5 +  .25=8.75  in.=length  of  brick  and  joint. 
2.375 +.25=  2.625  in.  =  thickness  of  brick  and  joint. 
8.75  x  2.625  =  22.96875  sq.  in.  =  area  of  its  face. 

12.75-^3  (number  of  bricks  in  width  of  wall)  =  4.25  m.=icidthoi  brick 
and  mortar. 

22.96875  x  4.25  =  97.617+  =  cubic  inches  in  a  brick. 
1728-^-97.617+  =  17.7+  =  number  of  bricks  in  a  cubic  foot. 

RULES.  I.  Add  to  the  face  dimensions  of  the  kind  of  bricks  used, 
one-half  the  thickness  of  the  mortar  or  cement  in  which  they  are  laid, 
and  compute  the  area. 

II.  Multiply  this  area  by  the  quotient  of  the  thickness  of  the  wall 
divided  by  the  number  of  bricks  of  which  it  is  composed,  the  product 
will  be  the  volume  of  a  brick  and  its  mortar  in  cubic  inches. 

III.  Divide  1728  by  this  volume,  and  the  quotient  will  be  the  num- 
ber of  bricks  in  a  cubic  foot. 

2.  How  many  bricks,  8  in.  x  4  x  2,  will  be  required  to  build  a 
wall  42  ft.  long,  24  ft.  high,  and  16-J  in.  thick,  laid  in  courses  of 
mortar  J  of  an  inch  thick?  Ans.   31278^. 

3.  How  many  perches  of  stone,  laid  dry,  will  build  a  wall  60  ft. 
long,  16  J  ft.  high,  and  18  in.  thick?  Ans.  60  Pch. 


450  MENSURATION. 

RULES.  1.  Multiply  the  number  of  cubic  feet  in  the  wall,  or  work 
to  be  done,  by  the  number  of  bricks  in  a  cubic  foot ;  the  product  will 
be  the  number  of  bricks  required. 

2.  Divide  the  number  of  cubic  feet  in  the  work  to  be  done  by  24.75  ; 
the  quotient  will  be  the  number  of  perches. 

4.  How  many  perches  of  masonry  in  a  wall  120  ft.  long,  6  ft. 
9  in.  high,  and  18  in.  thick? 

5.  At  $.56  a  cubic  yard,  what  will  it  cost  to  remove  an  embank- 
ment 240  ft.  long,  38  ft.  wide,  and  8.5  ft.  high? 

6.  Find  the  cost  of  digging  and  walling  the  cellar  of  a  house, 
whose  length  is  41  ft.  3  in.,  and  width  33  ft. ;  the  cellar  to  be  8  ft. 
deep,  and  the  wall  1-|-  ft.  thick.     The  excavating  will  cost  $.50  a 
load,  and  the  stone  and  mason  work  $3.75  a  perch.    Ans.  $47 If. 

7.  What  will  be  the  cost  of  building  a  wall  60  feet  long,  21-J  feet 
high,  and  17  inches  thick,  of  Philadelphia  bricks,  laid  in  courses  of 
mortar  J  of  an  inch  thick,  at  $12£  per  M.  ?  Ans.  $423.53. 

8.  How  many  cubic  feet  of  masonry  in  the  wall  of  a  cellar  37-J-feet 
long,  26  feet  wide,  and  9  feet  deep,  the  wall  being  2  feet  thick, 
allowing  one-half  for  the  corners ;  and  what  will  be  the  cost,  at 
$3.85  a  perch?  Ans.  2214  cu.  ft.;  $344.40. 

CAPACITY     OF     BINS,     CISTERNS,     ETC. 

794.  The  Standard  Bushel   of  the   United   States  contains 
2150.42  cu.  in.,  and  is  a  cylindrical  measure  18J  inches  in  diameter 
and  8  inches  deep  («511). 

1.  Measures  of  capacity  are  all  cubic,  measures,  solidity  and  capacity 
being  measured  by  different  nnits,  as  seen  in  the  tables. 

2.  Grain  is  shipped  from  New  York  by  the  Quarter  of  480  Ib.  (8  U.  S. 
bu.),  or  by  the  ton  of  33*  U.  S.  bushels. 

3.  It  is  sufficiently  accurate  in  practice  to  call  5  stricken  measures  equal 
to  4  heaped  measures. 

795.  To  find  the  exact  capacity  of  a  bin  in  bushels. 

RULES.  1.  Divide  the  contents  in  cubic  inches  by  2150.42  ;  the 
quotient  will  represent  the  number  of  bushels. 

Sincn  a  standard  bushel  contains  2150.42  cu.  in.,  and  a  cubic  foot  con- 
tains 1728  cu.  in.,  a  bushel  is  to  a  cubic  foot  nearly  as  4  to  5  ;  or  a  bushel 
is  equal  to  1]  cu.  ft.,  nearly.  Hence  for  all  practical  purposes, 


CAPACITY    OF    BINS,    CISTERNS,    ETC.  451 

2.  Any  number  of,  cubic  feet  diminished  by  -J  will  represent  an 
equivalent  number  of  bushels. 

Thus,  250  cu.  ft.  —  A  of  250  cu.  ft.,  or  50  cu.  ft.  =  200,  the  number  of 
bushels  in  250  cubic  feet 

3.  Any  number  of  bushels  increased  by  £  will  represent  an  equiva* 
lent  number  of  cubic  feet. 

Thus,  200  bu.  +  i  of  200  bu.,  or  50  bu.  =  250,  the  number  of  cubic  feet 
in  200  bushels. 

PROBLEMS. 

1.  How  many  bushels  of  wheat  can  be  put  in  a  bin  8  ft.  long, 
6  ft.  6  in.  wide,  and  3  ft.  4  in.  deep  ? 

2.  What  must  be  the  depth  of  a  bin  to  contain  240  bu.,  its  length 
being  10  feet  and  its  width  5  feet? 

OPERATION.— 240  bu.  +  60  bu.  =  300 ;  300-^10 x 5  =  6  ft.,  the  depth. 
RULE. — Divide  the  contents  in  cubic  feet  or  inches  by  the  product 
of  the  two  dimensions,  in  the  same  denomination. 

3.  What  must  be  the  length  of  a  bin  that  is  6  feet  wide  and  4-|  feet 
deep,  to  contain  324  bushels?  Ans.   15  ft. 

4.  How  many  bushels  of  apples  will  a  bir   hold,  that  is  12  ft. 
long,  3  ft.  wide,  and  2  ft.  6  in.  deep  ?     How  many  of  barley  ? 

5.  A  bin  20  ft.  long,  12  ft.  wide,  and  5  ft.  deep,  is  full  of  wheat. 
What  is  its  value  at  $2  a  bushel  ?  Ans.  $1920. 

6.  A  bin  7  ft.  long,  6  ft.  wide,  and  5  ft.  deep,  is  J  full  of  rye. 
What  is  its  value  at  $1.37£  a  bushel? 

7.  A  crib,  the  inside  dimensions  of  which  are  15  ft.  long,  7  ft. 

4  in.  wide,  and  8  ft.  high,  is  full  of  corn  in  the  ear.     If  2  bushels  of 
ears  make  1  bushel  of  shelled  corn,  what  is  the  value  of  the  whole, 
when  shelled,  at  -$.92  a  bushel  ?  Ans.  $259.07. 

8.  If  1  bu.  or  60  Ib.  of  wheat  make  48  Ib.  of  flour,  how  many 
barrels  of  flour  can  be  made  from  the  contents  of  a  bin  10  ft.  long, 

5  ft.  wide,  and  4  ft,  deep,  filled  with  wheat  ?         Ans.   39^  bbl. 

9.  Dunkley  &  Co.  bought  12400  bu.  of  wheat,  delivered  in  New 
Y"ork,  at  $1,50  a  bushel.      They  shipped  the  same  to  Liverpool, 
paying  6s.  sterling  per  quarter  freight,  and  sold  the  entire  cargo  at 
12s.  per  cental.     Making  no  allowance  for  exchange  or  for  waste, 
what  was  the  gross  gain  in  U.S.  Money,  the  £  being  valued  at  &4.666J-? 


452  MENSUKATIOIST. 

796.  To  find  the  exact  capacity  of  a  vessel  or  space  in 
gallons. 


PROBLEMS. 


1.  How  many  gallons  of  water  will  a  cistern  hold,  that  is  4  feet 
square  and  6  feet  deep  ? 

OPERATION.— (4  x  4  x  6  x  172§)-^231  -  718|f  gal.,  capacity. 

RULE.     Divide  the  contents  in  cubic  inches  by  231  for  liquid  gal- 
lons, or  by  2  6  8. 8  for  dry  gallons. 

2.  How  many  cubic  feet  in  a  space  that  holds  48  hhd.  ? 

3.  How  many  hogsheads  will  a  cistern   11  ft.  long,  6  ft.  wide, 
and  7  feet  deep  contain  ?  Ans.   54&  hhd. 

4.  How  many  gallons  will  a  space  contain  that  is  22.5  ft.  long, 
3.25  ft.  wide,  and  6.4  ft.  deep? 

5.  A  man  constructed  a  cistern  to  hold  32  hogsheads,  the  bottom 
being  6  ft.  by  8  ft.     What  was  its  depth  ?          Ans.  5  ft.  7f  in. 

6.  A  tank  in  the  attic  of  a  house  is  6  ft.  6  in.  long,  4  ft.  wide, 
and  3  ft.  6  in.  deep.     How  many  gallons  of  water  will  it  hold,  and 
what  will  be  its  weight?  Ans.  680-^-  gal. ;  5672^r  Ib. 

7.  If  64  quarts  of  water  be  put  into  a  vessel  that  will  exactly 
hold  64  quarts  of  wheat,  how  much  will  the  vessel  lack  of  being 
full?  Ans.  604.8  cu.  in. 

8.  If  a  man  buy  10  bu.  of  chestnuts  at  $5  a  bushel,  dry  measure, 
and  sell  the  same  at  25  cents  a  quart,  liquid  measure,  how  much 
does  he  gain?  Ans.  $43.09. 

9.  A  cistern   5  ft.  by  4  ft.  by  3  ft.  is  full  of  water.     If  it  be 
emptied  by  a  pipe  in  1  hr.  30  min.,  how  many  gallons  are  discharged 
through  the  pipe  in  a  minute  ?  Ans.  4-J-f-  gal. 

10.  A  vat  that  will  hold  5000  gallons  of  water,  will  hold  how 
many  bushels  of  corn  ?  Ans.  537-^  bu. 

11.  A  cellar  40  ft.  long,  20  ft.  wide,  and  8  ft.  deep  is  half-full 
of  water.     What  will  be  the  cost  of  pumping  it  out,  at  6  cents  a 
hogshead?  Ans.  $22.80. 

12.  A  reservoir  24  ft.  8  in.  long  by  12  ft.  9  in.  wide  is  full  oi 
water.     How  many  cubic  feet  must  be  drawn  off  to  sink  the  sur* 
face  1  foot?     How  many  gallons?  Ans.  2352^-f  gal. 


THE    METEIC    SYSTEM 
OF     WEIGHTS    AND    MEASURES. 

797.  The  Metric  System  was  adopted  in  France  in  1795;  its 
use  was  authorized  in  Great  Britain  in  1864;  and  in  1866,  Con- 
gress authorized  the  Metric  System  to  be  used  in  the  United  States 
by  passing  the  following  bills : 

AN   ACT    TO   AUTHORIZE   THE   USE    OF   THE   METRIC    SYSTEM   OP 
WEIGHTS  AND  MEASURES. 

Be  it  enacted  by  the  Senate  and  House  of  Representatives  of  the  United 
States  of  America  in  Congress  assembled,  That  from  and  after  the  passage 
of  this  Act,  it  shall  be  lawful  throughout  the  United  States  of  America 
to  employ  the  Weights  and  Measures  of  the  Metric  System  ;  and  no  con- 
tract or  dealing,  or  pleading  in  any  court,  shall  be  deemed  iuvalid,  or 
liable  to  objection,  because  the  weights  or  measures  expressed  or  referred 
to  therein  are  weights  or  measures  of  the  Metric  System. 

SECTION  2.  And  be  it  further  enacted,  That  the  tables  in  the  schedule 
hereto  annexed  shall  be  recognized  in  the  construction  of  contracts,  and 
in  all  legal  proceedings,  as  establishing,  in  terms  of  the  weights  and 
measures  now  in  use  in  the  United  States,  the  equivalents  of  the  weights 
and  measures  expressed  therein  in  terms  of  the  Metric  System  ;  and  said 
tables  may  bs  lawfully  used  for  computing,  determing,  and  expressing 
in  customary  weights  and  measures,  the  weights  and  measures  of  the 
Metric  System. 

708.  The  Metric  System  of  weights  and  measures  is  based 
upon  the  decimal  scale. 

700.  The  Meter  is  the  base  of  the  system,  and  is  the  one  ten* 
millionth  part  of  the  distance  on  the  earth's  surface  from  the  equa- 
tor  to  either  pole,  or  39.37079  inches. 

8O©.  From  the  Meter  are  made  the  Are  (air),  the  Stere  (stair), 
the  Liter  (leeter),  and  the  Gram  ;  these  constitute  the  primary  or 
principal  units  of  the  system,  from  which  all  the^ others  are  derived. 

8O1.  The  Multiple  Units,  or  higher  denominations,  are  named 
by  prefixing  to  the  name  of  the  primary  units  the  Greek  numerals, 
Deka  (10),  Hecto  (100),  Kilo  (1000),  and  Myra  (10000). 


454  MENSUKATION. 


The  Sub-multiple  Units,  or  lower  denominations,  are 
named  by  prefixing  to  the  names  of  the  primary  units  the  Latin 
numerals,  Deci  (fa),  Centi  (^),  Milk  (T^). 

Hence,  it  is  apparent  from   the  name  of  a  unit,  whether  it  is 
greater  or  less  than  the  standard  unit,  and  also  how  many  times. 

MEASURES    OF   EXTENSION. 

§®8o   The  Meter  is  the  unit  of  length,  and  is  equal  to  39.37  in. 
nearly. 

TABLE. 


Metric  Denominations. 

U.  S.  Value. 

1  Millimeter 

=       .03937079  in. 

10  Millimeters,     mm. 

=  1  Centimeter 

=       .3937079  in. 

10  Centimeters,    cm. 

=  1  Decimeter 

=     3.937079  in. 

10  Decimeters,     dm. 

=  1  METER 

-  39.37079  in. 

10  METERS          M. 

=  1  Dekameter 

=  32.808992  ft. 

10  Dekameters,   Dm. 

—  1  Hectometer 

=  19.927817  rd. 

10  Hectometers,  Hm. 

=  1  Kilometer 

=       .6213824mi. 

10  Kilometers,     Km. 

—  1  Myrianieter  (Mm.) 

=     6.213824  mi. 

Tlie  meter,  like  our  yard,  is  used  in  measuring  cloths  and  short  dis- 
tances. 

The  kilometer  is  commonly  used  for  measuring  long  distances,  and 
is  about  of  a  common  mile. 


The  Are  is  the  unit  of  land  measure,  and  is  a  square 
whoso  side  is  10  meters,  equal  to  a  square  dekameter,  or  119.6  sq. 
yards. 

TABLE. 

1  Centiare,   ca.     =  (1  Sq.  Meter)       =      1.196034  sq.  yd. 
100  Centiares,  "      =  1  ARE  =  119.  6034  sq.  yd. 

100  ARES         A.     =  1  Hectare  (Ha.)  =      2.47114  acres. 


The  Square  Meter  is  the  unit  for  measuring  ordinary 
surfaces  ;  as  flooring,  ceilings,  etc. 


TABLE. 


100  Sq.  Millimeters,  sq.mm.  =  1  Sq.  Centimeter        =      .155+  sq.in. 
100  Sq.  Centimeters,  sq.cm.   =  1  Sq.  Decimeter          =  15.5+  sq.  in. 
100  Sq.  Decimeters,  sq.  dm.  =  1  SQ.  METEE  (8q.M.)  =  1.196+  sq.  yd 


METBIC    SYSTEM.  455 

The  Stere  is  the  unit  of  wood  or  solid  measure,  and  is 
equal  to  a  cubic  meter,  or  .2759  cord. 

TABLE. 

1  Decistere  =    3.531+  cu.  ft. 

10  Decisteres,  dst.  =  1  STERE  =  35.316  +  cu.  ft. 

10  STEKES       8t.    =  1  Dekastere  (DSt.)  =  13.079+  cu.  yd. 

807.  The  Cubic  Meter  is  the  unit  for  measuring   ordinary 
solids;  as  excavations,  embankments,  etc. 

TABLE. 

1000  Cu.  Millimeters,  cu.  mm.  =  1  Cu.  Centimeter  =      .001  +  cu.  in. 
1000  Cu.  Centimeters,  CM.  cm.    =  1  Cu.  Decimeter    =  61.026  +    "    " 
1000  Cu.  Decimeters,  cu.  dm.  =  1  Cu.  METER        =  35.316  +  cu.  ft. 

MEASURES    OF   CAPACITY, 

808.  The  Liter  is  the  unit  of  capacity,  both  of  Liquid  and  of 
Dry  Measures,  and  is  a  vessel  whose  volume  is  equal  to  a  cube 
whose  edge  is  one-tenth  of  a  meter,  equal  to  1.G5673  qt.  Liquid 
Measure,  and  .9081  qt.  Dry  Measure, 

TABLE. 

10  Milliliters,     ml.    .     .     .  =  1  Centiliter. 

10  Centiliters,    cl.      .     .     .  —  1  Deciliter.  w 

10  Deciliters,     ctt.     .     ...  =1  LITER. 

10  LITERS         L.     ...  =1  Dekaliter.- 

10  Dekaliters,   Dl.    .     .     .  =  1  Hectoliter. 

10  Hectoliters,  El.    .     .     .  =  1  Kilcliter,  or  Stere. 

10  Kiloliters,     El.    .     .     .  =  I  Myrialiter  (Ml.). 
The  Hectoliter  is  the  unit  in  measuring  liquids,  grain,  fruit,  and  roots 
in  large  quantities. 

8 SO.         EQUIVALENTS    IX    UNITED    STATES    MEASURES. 

Metric  Denominations.    Cubic  Measure.  Dry  Measure.  Wine  Measure. 

1  Myrialiter  =   10  Cubic  Meters       =  283.72+  bu.  =  2641.4+  gal. 

1  Kilpliter  =     1  Cubic  Meter        =  28.372+  bu.  -  264.17  gal. 

1  Hectoliter  —  -^  Cubic  Meter         =  2.8372+  bu.  =  26.417  gal. 

1  Dekaliter  =   10  Cu.  Decimeters   =  9.08  quarts  =  2.6417  gal. 

1  LITER  —     1  Cu.  Decimeter     =  .908  quart  =  1.0567  qt. 

1  Deciliter  =    ^  Cu.  Decimeter     =  6.1022  cu.  in.  =  .845  gill. 

1  Centileter  =  10  Cu.  Centimeters  =  .6102  cu.  in.  =  .338  fluid  oz. 

1  MUliliter  =     1  Cu.  Centimeter  =  .061  cu.  in.  =  .27  fluid  dr. 


456 


MENSURATION. 


MEASURES    OF   WEIGHT. 

8 1 0.  The  Gram  is  the  unit  of  loeight,  and  equal  to  the  weight 
of  a  cube  of  distilled  water,  the  edge  of  which  is  one-hundredth  of 
a  meter,  equal  to  15.432  Troy  grains. 


TABLE. 


10  Milligrams, 
10  Centigrams, 
10  Decigrams, 
10  GRAMS 
10  Dekagrams, 

10  Hectograms, 

mg.    —  1  Centigram 
eg.      =  1  Decigram 
dg.     =  1  GRAM 
G.      =1  Dekagram 
Dg.    =  1  Hectogram 

rj-          ^  (  Kilogram, 
Jig.    —  L  -i 

.15432+  gr.  Troy. 
1.54324+  "      " 
15.43248+  "       " 

.35273+  oz.  Avoir. 
3.52739+  "      " 


10  Kilograms,          Kg. 

10  Myriagrams,  oi,Mg.  ) 

100  Kilograms  ) 

10  Quintals,  or  ) 

1000  KILOS  ) 


=  1 


1  Myriagram      =     22.04621+  " 
1  Quintal  =  220.46212+  " 

( Tonneau, 


(    or  TON 


j-  =2204.62125 


The  Kilogram,  or  Kilo,  is  the  unit  of  common  weight  in  trade  and  is  a 
trifle  less  than  2^  Ib.  Avoirdupois. 

The  Tonneau  is  used  for  weighing  very  heavy  articles,  and  is  abouT 
204  Ib.  more  than  a  common  ton. 


811.  Units  of  the  Common  System  maybe  readily  changed 
to  units  of  the  Metric  System  by  the  aid  of  the  following 


TABLE. 


1  Inch 
1  Foot        : 
1  Yard       ; 
1  Rod         : 
1  Mile 
1  Sq.  inch  ; 
1  Sq.  foot 
1  Sq.  yard 
1  Sq.  rod 
1  Acre 
1  Sq.  mile 


2.54  Centimeters. 
30.48  Centimeters. 
.9144  Meter. 
5.029  Meters. 
1.6093  Kilometers. 
6.4528  Sq.  Centimet. 
929  Sq.  Centimeters. 
.8361  Sq.  Meter. 
25.29  Centiares. 
40.47  Ares. 
259  Hectares. 


i  -t  9* 


Cu.  inch 

Cu.  foot  : 

Cu.  yard : 

Cord        : 

Fl.  ounce: 

Gallon     : 

Bushel    ; 

1  Troy  gr.  : 

1  Troy  Ib.  : 

1  Av.  Ib.    : 

1  Ton         ; 


:  16.39  Cu.  Centimet. 
28320  Cu.  Centimet. 
.7646  Cu.  Meter. 
3.625  Steres. 
2.958  Centiliters. 
3.786  Liters. 
.3524  Hectoliter. 
64.8  Milligrams. 
.373  Kilo. 
.4530  Kilo. 
.907  Tonneau. 


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